一些因吹斯汀的题目记录

思维
贪心
- 可反悔贪心
- 模拟退火
并查集
dp
分治法
- 最近点对
dfs
组合数学——隔板法
离散化 + 树状数组

思维

题：http://lx.lanqiao.cn/problem.page?gpid=T2908
题意：给定序列（a1，a2，……an），进行 m 次操作，p = 0，a1 ~ aq 降序排序，p = 1，aq ~ an 升序排序。
思路：如果是连续相同的操作，则取最大区间，可得两个操作是间或的，需要翻转的区间越来越小。根据操作去把数放好。

#include <iostream>#define x first
#define y secondusing namespace std;typedef pair<int, int> PII;const int N = 1e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2 int n, m;
PII stk[N];
int ans[N];signed main()
{scanf("%d%d", &n, &m);int top = 0;while (m--){int p, q;scanf("%d%d", &p, &q);if (!p){while (top && stk[top].x == 0)q = max(q, stk[top--].y);while (top >= 2 && stk[top - 1].y <= q)top -= 2;stk[++top] = {0, q};}elseif (top){while (top && stk[top].x == 1)q = min(q, stk[top--].y);while (top >= 2 && stk[top - 1].y >= q)top -= 2;stk[++top] = {1, q};}}int k = n, l = 1, r = n;for (int i = 1; i <= top; ++i){if (stk[i].x == 0)while (r > stk[i].y && l <= r)ans[r--] = k--;elsewhile (l < stk[i].y && l <= r)ans[l++] = k--;if (l > r)break;}if (top % 2)while (l <= r)ans[l++] = k--;elsewhile (l <= r)ans[r--] = k--;for (int i = 1; i <= n; ++i)printf("%d ", ans[i]);return 0;
}

贪心

可反悔贪心

题：https://www.luogu.com.cn/problem/P1484

采用双向链表实现。

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
//#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
//#include <random>
#include <map>
//#include <unordered_map>    #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long
#define endl '\n' #define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef pair<LL, int> PLI;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 5e5+ 5, MOD = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2
const double pi = 3.141592653589;double func(double x, double y)   //  评估函数
{return 6 * pow(x, 7.0) + 8 * pow(x, 6.0) + 7 * pow(x, 3.0) + 5 * pow(x, 2.0) - y * x;
}LL gcd(LL a, LL b)
{return b == 0 ? a : gcd(b, a % b);
}LL lcm(LL a, LL b)
{return a / gcd(a, b) * b;
}LL exgcd(LL a, LL b, LL &x, LL &y)
{if (!b){x = 1, y = 0;return a;}LL d = exgcd(b, a % b, y, x);y -= (a / b) * x;return d;
}LL wow[N];
int pre[N], nex[N];
bool v[N];
priority_queue<PLI> q;inline void del(int u)
{nex[pre[u]] = nex[u];pre[nex[u]] = pre[u];
}signed main()
{IOS;int n, m;cin >> n >> m;for (int i = 1; i <= n; ++i){cin >> wow[i];q.push({wow[i], i});pre[i] = i - 1;nex[i] = i + 1;}LL ans = 0, res = 0;while (m--){auto t = q.top();q.pop();while (v[t.second]){t = q.top();q.pop();}int u = t.second;v[nex[u]] = v[pre[u]] = true;ans += wow[u];if (res >= ans)break;elseres = ans;wow[u] = wow[nex[u]] + wow[pre[u]] - wow[u];del(nex[u]);del(pre[u]);q.push({wow[u], u});}cout << res << endl;return 0;
}

题：https://ac.nowcoder.com/acm/contest/32942/A

模拟退火

题：https://vjudge.net/contest/488791#problem/H
题意： F(x) = 6 * x^7 + 8 * x^6 + 7 * x^3 + 5 * x^2 - y * x (0 <= x <=100)。给出y的值，求F(x)的最小值。

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
#include <random>
#include <map>
#include <unordered_map>  #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
#define int long long
#define endl '\n' #define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 2e6 + 5, MOD = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2
const double pi = 3.141592653589;const double T0 = 1e5, TK = 1e-8, d = 0.98;    //初始温度为高温，设置成一个大数；终止温度为低温，设置为一个接近于0的数；d是一个小于1但是非常接近于1的数double func(double x, double y)  //  评估函数
{return 6 * pow(x, 7.0) + 8 * pow(x, 6.0) + 7 * pow(x, 3.0) + 5 * pow(x, 2.0) - y * x;
}double simulateAnneal(double y)
{double T = T0, x = 50.0; //初始值double now = func(x, y);  //当前状态double nxt;   //新状态double ans = now; //最优解int f[2] = {-1, 1};while (T > TK){double newx = x + f[rand() % 2] * T;   //按概率改变x，温度越低概率越小if (newx >= 0 && newx <= 100){nxt = func(newx, y); //新状态ans = min(ans, nxt);  //在退火过程中维护遇到的所有解的最优值double dE = now - nxt; //能量差if (dE >= 0)   //新状态更优就接受{now = nxt;x = newx;}elseif (exp(dE / T) > rand())   //以一定概率接受{now = nxt;x = newx;}}T *= d;   //降温}return ans;
}signed main()
{IOS;int T;cin >> T;while (T--){double y;cin >> y;cout << fixed << setprecision(4) << simulateAnneal(y) << endl;}return 0;
}

并查集

题：https://vjudge.net/problem/POJ-1182
题意：A吃B，B吃C，C吃A。有两个操作，op=1，x和y是一个物种，op=2，x吃y，根据前面的真话去判断当前的话是真是假，输出假话数量。
思路：根据维护到祖宗结点距离的并查集，可以知道，有三层集合，0、1、2。1吃0，2吃1，3吃2，由于这是一个食物环，所以3跟0是一类的，可以根据距离%3判断是哪个集合的，然后判断当前的话是真是假。

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
//#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
//#include <random>
#include <map>
//#include <unordered_map>    #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long
#define endl '\n' #define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 5e4 + 5, MOD = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2
const double pi = 3.141592653589;double func(double x, double y)   //  评估函数
{return 6 * pow(x, 7.0) + 8 * pow(x, 6.0) + 7 * pow(x, 3.0) + 5 * pow(x, 2.0) - y * x;
}int n, m;
int p[N], d[N];int find(int x)
{if (p[x] != x){int t = find(p[x]);d[x] += d[p[x]];p[x] = t;}return p[x];
}signed main()
{scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i)p[i] = i;int ans = 0;while (m--){int op, x, y;scanf("%d%d%d", &op, &x, &y);if (x > n || y > n)ans++;else{int px = find(x), py = find(y);if (op == 1){if (px == py && (d[x] - d[y]) % 3)ans++;elseif (px != py){p[px] = py;d[px] = d[y] - d[x];}}else{if (px == py && (d[x] - d[y] - 1) % 3)ans++;elseif (px != py){p[px] = py;d[px] = d[y] + 1 - d[x];}}}}printf("%d\n", ans);return 0;
}

dp

题：https://ac.nowcoder.com/acm/contest/32907/I

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
//#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
//#include <random>
#include <map>
//#include <unordered_map>    #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long
#define endl '\n' #define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 5e4 + 5, MOD = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2
const double pi = 3.141592653589;double func(double x, double y)   //  评估函数
{return 6 * pow(x, 7.0) + 8 * pow(x, 6.0) + 7 * pow(x, 3.0) + 5 * pow(x, 2.0) - y * x;
}LL gcd(LL a, LL b)
{return b == 0 ? a : gcd(b, a % b);
}LL lcm(LL a, LL b)
{return a / gcd(a, b) * b;
}LL exgcd(LL a, LL b, LL &x, LL &y)
{if (!b){x = 1, y = 0;return a;}LL d = exgcd(b, a % b, y, x);y -= (a / b) * x;return d;
}LL n, t, dp[105][5205][105], w[105], v[105];signed main()
{scanf("%lld%lld", &n, &t);for (int i = 1; i <= n; ++i)scanf("%lld%lld", &w[i], &v[i]);//这里用memset会超时for (int i = 0; i <= n; ++i)for (int j = 0; j <= 5200; ++j)for (int k = 0; k <= t; ++k)dp[i][j][k] = -INF;for (int k = 0; k <= t; ++k) //把n=1的情况单独更新出来{//选第一张卡if (k){dp[1][2600 + 2 * v[1]][k] = w[1];dp[1][2600 - 2 * v[1]][k] = w[1];}dp[1][2600 - v[1]][k] = w[1];dp[1][2600 + v[1]][k] = w[1];//不选第一张卡dp[1][2600][k] = 0;}for (int i = 2; i <= n; ++i)for (int j = 0; j <= 5200; ++j)for (int k = 0; k <= t; ++k){//不翻倍且不把第i张卡放入任何集合dp[i][j][k] = dp[i - 1][j][k];//翻倍且不把第i张卡放入任何集合的情况，不优，舍掉//dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j][k - 1]);//把第i张卡翻倍放入A集合if (k >= 1 && j - 2 * v[i] >= 0)dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j - 2 * v[i]][k] + w[i]);//把第i张卡不翻倍放入A集合if (j - v[i] >= 0)dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j - v[i]][k] + w[i]);//把第i张卡翻倍放入B集合if (k >= 1 && j + 2 * v[i] <= 5200)dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j + 2 * v[i]][k - 1] + w[i]);//把第i张卡不翻倍放入B集合if (j + v[i] <= 5200)dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j + v[i]][k] + w[i]);}printf("%lld\n", dp[n][2600][t]);return 0;
}

题：https://codeforces.com/contest/255/problem/C
题意：找到一个最长子序列p,q,p, q。（可以在给出的序列中任意删除元素）

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>  #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long#define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 4e3 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2 int wow[N], dp[N][N];signed main()
{   int n, ans = 0;scanf("%d", &n);for (int i = 1; i <= n; ++i)   //子序列最后一个数是 i {scanf("%d", &wow[i]);for (int j = 0, id = 0; j < i; ++j)    //子序列倒数第二个数为 j{ dp[i][j] = dp[j][id] + 1;if (wow[i] == wow[j])id = j;  // id 记录前一个相等的数的位置ans = max(ans, dp[i][j]);}}printf("%d\n", ans);return 0;
}

题：https://ac.nowcoder.com/acm/contest/30789/D
题意：给出N个数，问是否能从中选出一些数字，使得其加和为3600的倍数

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>  #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long#define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 1e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2 int wow;
bool dp[N], cnt[N];
int n;signed main()
{   int T;scanf("%d", &T);while (T--){memset(dp, false, sizeof dp);memset(cnt, false, sizeof cnt);int n;scanf("%d", &n);for (int i = 1; i <= n; ++i){scanf("%d", &wow);wow %= 3600;if (!dp[0]){for (int j = 0; j < 3600; ++j)if (dp[j] || !j)cnt[(wow + j) % 3600] = true;    //跟当前每个存在的数组成新值录入for (int j = 0; j < 3600; ++j)   //从cnt转存进dp{if (cnt[j])dp[j] = true;if (cnt[j])cnt[j] = false;}}}if (!dp[0])  //判断是否能组成模为0的printf("NO\n");elseprintf("YES\n");}return 0;
}

题：http://lx.lanqiao.cn/problem.page?gpid=T2900
题意：给定一个括号序列，要求尽可能少添加若干括号使之合法，求方案数。mod = 1e9 + 7。
思路：
左括号和右括号的插入是独立的，可以直接方案相乘起来。
插入左括号和插入右括号本质相同，将序列翻转后，每个字符也翻转即可。

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;typedef long long LL;const int N = 5010, MOD = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2 int n;
char str[N];
LL f[N][N];LL calc()
{memset(f, 0, sizeof f);f[0][0] = 1;for (int i = 1; i <= n; ++i)if (str[i] == '(')for (int j = 1; j <= n; ++j)f[i][j] = f[i - 1][j - 1];else{f[i][0] = (f[i - 1][0] + f[i - 1][1]) % MOD;for (int j = 1; j <= n; ++j)f[i][j] = (f[i - 1][j + 1] + f[i][j - 1]) % MOD;}for (int i = 0; i <= n; ++i)if (f[n][i])return f[n][i];return -1;
}signed main()
{scanf("%s", str + 1);n = strlen(str + 1);LL l = calc();reverse(str + 1, str + n + 1);for (int i = 1; i <= n; ++i)if (str[i] == '(')str[i] = ')';elsestr[i] = '(';LL r = calc();printf("%lld\n", l * r % MOD);return 0;
}

题：https://ac.nowcoder.com/acm/contest/11224/D
题意：按顺序选一些字符串拼接起来，求字符串最大和，且两个拼接的字符串中前者的尾要与后者的头相同。
思路：

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>  #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long
#define endl '\n' #define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 2e5 + 10, MOD = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2 int f[27];signed main()
{IOS;int T;cin >> T;while (T--){memset(f, 0, sizeof f);int n;cin >> n;for (int i = 1; i <= n; ++i){string s;cin >> s;f[s.back() - 'a'] = max(f[s.back() - 'a'], (int)s.size() + f[s[0] - 'a']);}int res = 0;for (int i = 0; i < 26; ++i)res = max(res, f[i]);cout << res << endl;}return 0;
}

题：https://vjudge.net/contest/488040#problem/F

分治法

dfs

题：https://atcoder.jp/contests/abc236/tasks/abc236_d
题意：共有2N个人，编号为1-2N，现要人们两两配对，若编号为i的人想与编号为j的人配对，i必须小于j，i与j配对后获得亲和力B。所有人配对完成后亲和力分别为B1,B2…BN，总乐趣为B1⊕B2⊕…⊕BN。求所有配对组合中总乐趣的最大值。

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>  #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long
#define endl '\n' #define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 3e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2 int n, m, ans = 0;
vector <int> t;
int wow[20][20];
bool vis[20];void dfs(int p)
{if (t.size() == n){int tmp = t[0];for (int i = 1; i < t.size(); ++i)tmp ^= t[i];ans = max(ans, tmp);return ;}if (vis[p]){dfs(p + 1);return ;}for (int i = p + 1; i < 2 * n; ++i){if (!vis[i]){vis[p] = vis[i] = true;t.push_back(wow[p][i]);dfs(p + 1);t.pop_back();vis[p] = vis[i] = false;}}
}signed main()
{IOS;cin >> n;m = 2 * n;for (int i = 0, x; i < m - 1; ++i)for (int j = i + 1; j < m; ++j)cin >> wow[i][j];dfs(0);cout << ans << endl;return 0;
}

题：https://codeforces.com/problemset/problem/616/C
题意：给出n * m大小的只有‘ * ’和‘ . ’的矩阵，分别计算每一个’ * ‘的连通’ . ‘，当计算该’ * ‘时，可将该’ * ‘视为’ . ‘。
思路：首先预处理一下，每个’ . ‘的连通块计算出有多少并且标记，然后再分别处理每个’ * '四个方向的标记，如果该标记未出现过则加进ans。

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <set>
#include <bitset>
#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>  #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long
#define endl '\n' #define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 1e3 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
const double eps = 1e-8;   //一般比小数点的位数至少多 2 int n, m, tmp, cnt = 0;
char wow[N][N];
bool vis[N][N];
int hh[N][N], num[N * N];
map <int, int> mp;void dfs(int x, int y)
{hh[x][y] = cnt;vis[x][y] = 1;for (int i = 0; i < 4; ++i)if (x + dx[i] >= 0 && x + dx[i] < n && y + dy[i] >= 0 && y + dy[i] < m && wow[x + dx[i]][y + dy[i]] == '.' && !vis[x + dx[i]][y + dy[i]]){tmp++;dfs(x + dx[i], y + dy[i]);}
}signed main()
{scanf("%d%d", &n, &m);for (int i = 0; i < n; ++i)scanf("%s", wow[i]);for (int i = 0; i < n; ++i)for (int j = 0; j < m; ++j)if (wow[i][j] == '.' && !hh[i][j]){tmp = 1;cnt++;dfs(i, j);num[cnt] = tmp;}for (int i = 0; i < n; ++i){for (int j = 0; j < m; ++j)if (wow[i][j] == '*'){int ans = 0;for (int k = 0; k < 4; ++k)if (i + dx[k] >= 0 && i + dx[k] < n && j + dy[k] >= 0 && j + dy[k] < m && hh[i + dx[k]][j + dy[k]]){if (!mp[hh[i + dx[k]][j + dy[k]]]){ans += num[hh[i + dx[k]][j + dy[k]]];mp[hh[i + dx[k]][j + dy[k]]]++;}}printf("%d", (ans + 1) % 10);mp.clear();}elseprintf(".");puts("");}   return 0;
}

组合数学——隔板法

题：https://ac.nowcoder.com/acm/contest/30184/L
题意：给出每个数在 1~1000 之间的单调不增序列，其中有一些数字丢失了，找出有多少方案满足该序列要求，答案对1000000007取模。

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <algorithm>
#include <set>
#include <bitset>
#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>  #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
//#define int long long#define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}const int N = 1e6 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;int a[N];
LL f[N];void init() //预处理
{f[0] = f[1] = 1;for (int i = 2; i <= 1e6; ++i)f[i] = (f[i - 1] * i) % mod;
}LL C(int n, int m) //排列组合
{if (m == 0)return 1;return (f[n] * fastpow(f[n - m] * f[m] % mod, mod - 2, mod) % mod) % mod;    //乘法逆元
}signed main()
{   init();int n;scanf("%d", &n);for (int i = 1; i <= n; ++i)scanf("%d", &a[i]);int cnt = 0, id = 0;LL ans = 1;a[0] = 1000;for (int i = 1; i <= n; ++i){if (!a[i])cnt++;    //小球else{int n = cnt;int m = a[id] - a[i] + 1;   //盒子id = i;cnt = 0;ans = ans * (C(n + m - 1, m - 1) % mod) % mod;}          }if (cnt)   //末尾数处理{int n = cnt, m = a[id];ans = ans * C(n + m - 1, m - 1) % mod;}printf("%lld\n", ans % mod);return 0;
}

离散化 + 树状数组

题：https://www.acwing.com/problem/content/4319/
题意：

#include <iostream>
#include <cstdio>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <algorithm>
#include <set>
#include <bitset>
#include <unordered_set>
#include <memory.h>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>  #define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define x first
#define y second
#define all(a) a.begin(), a.end()
#define int long long#define bug(a) cout << a << " *bug1*" << endl
#define bugg(a, b) cout << a << " " << b << " *bug2*" << endl
#define buggg(a, b, c) cout << a << " " << b << " " << c << " *bug3*" << endlusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};LL fastpow(LL b, LL k, LL p)
{LL res = 1;while (k){if (k & 1)res = res * 1LL * b % p;b = b * 1LL * b % p;k >>= 1;}//bug(res);return res % p;
}int lowbit(int x)
{return x & -x;
}const int N = 1e6 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;int n, m;
int s[N];
int tr[N];
vector <int> nums;void add(int x, int k)  //树状数组修改操作
{for (int i = x; i < N; i += lowbit(i))tr[i] += k;
}int sum(int x) //树状数组查询操作
{int res = 0;for (int i = x; i; i -= lowbit(i))res += tr[i];return res;
}int get(int x) //二分获取离散化之后的下标
{return lower_bound(all(nums), x) - nums.begin() + 1;
}signed main()
{   scanf("%lld%lld", &n, &m);for (int i = 1; i <= n; ++i){scanf("%lld", &s[i]);s[i] += s[i - 1];  //前缀和}for (int i = 0; i <= n; ++i)   //将所有前缀和以及后续询问用到的数值全部离散化{nums.push_back(s[i]);nums.push_back(s[i] - m);}sort(all(nums));nums.erase(unique(all(nums)), nums.end());int res = 0;add(get(s[0]), 1);   //将 0 这个位置的前缀和加入树状数组for (int i = 1; i <= n; ++i){//sum(1000000) 整个值域上所有值的总个数 //sum(get(s[i]−m)) 表示所有 i 之前的前缀和中小于等于 get(s[i]−m) 的前缀和的个数res += sum(1e6) - sum(get(s[i] - m));    //答案累加上树状数组中大于 sum[r] - t 的总个数add(get(s[i]), 1);    // sum[r] 这个位置的个数加上 1 ，方便后续的计算//枚举右端点，右端点不减1，左端点减1// sum(r) - sum(l-1) < t 即 sum(l-1) > sum(r) - t       }printf("%lld\n", res);return 0;
}