第九届河南理工大学算法程序设计大赛正式赛（ABCDEFGHJKL）

ACcode

A. Asia区域赛

读题目输出奖牌之和就行，我把冠亚军也记成奖牌了。。

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 1001;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;
const int mod = 1e9 + 7;int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);   int ans = 29 + 87 + 58 + 11 + 1 + 1;cout << ans << endl;return 0;
}

B. Asia区域制

二进制转化为十进制，当成字符串搞搞 “4位一并”

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 1001;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;
const int mod = 1e9 + 7;int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);   int T;cin >> T;map<int, char> mp;mp[10] = 'a';mp[11] = 'b';mp[12] = 'c';mp[13] = 'd';mp[14] = 'e';mp[15] = 'f';while (T--) {string s;cin >> s;string t;int len = s.size();int last = len % 4;for (int i = 0; last && i < 4 - last; ++i) {s = "0" + s;}len = s.size();int tmp = 0;int cnt = 0;for (int i = 0; i < len; ++i) {++tmp;cnt = cnt * 2 + s[i] - '0';if (tmp % 4 == 0) {// cout << tmp << endl;if (cnt <= 9) t += '0' + cnt;else t += mp[cnt];cnt = 0;}}cout << t << endl;}return 0;
}

C. Asia区域宫

每行每列只有一个障碍，只有存在一条对角线才可能把起点封住，读入点的时候预处理每种对角线的点数，如果存在点数合理就是NO，否者就是曼哈顿距离

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 10001;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;int n, m;
int sum[maxn];int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);  int T;cin >> T;while (T--) {mem(sum, 0);cin >> n >> m;for (int i = 0, x, y; i < m; ++i) {cin >> x >> y;sum[x+y-1]++;}int flag = 1;for (int i = 1; i <= n; ++i) {if (sum[i] == i) flag = 0;}if (flag) cout << "Yes " << 2*n - 2 << endl;else cout << "No\n";}return 0;
}

D. Asia区域阵

当时读题的时候感觉好难，数据给的很小，但是暴力的话感觉很难写，于是就放弃了。
补个暴力，枚举矩阵的起点然后计算以该点能扩展的最大矩阵，列扩展的时候需要标记，因为之后的列一定小于等于当前所能扩展的最小列

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 1000001;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;char a[101][101];
int R[101][91], C[101][91];
int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);  int T;scanf("%d", &T);while (T--) {int n, m;scanf("%d %d", &n, &m);for (int i = 0; i < n; ++i) scanf("%s", a[i]);int ans = 0;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {mem(R, 0);mem(C, 0);int c = m;for (int p = i; p < n; ++p) {for (int k = j; k < c; ++k) {int t = a[p][k];if (R[p][t] || C[k][t]) {c = k;break;}ans = max(ans, (p-i+1)*(k-j+1));R[p][t] = 1;C[k][t] = 1;}}}}printf("%d\n", ans);}return 0;
}

E. Mo的游戏

统计每个字符出现的位置，然后for一遍，找最小的差（最小的差一定相邻）

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;vector<int> g[1000];int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);  string s;cin >> s;int len = s.size();for (int i = 0; i < len; ++i) {int t = s[i];g[t].push_back(i);}for (int i = 'a'; i <= 'z'; ++i) {if (g[i].empty()) continue;int ll = g[i].size();if (ll == 1) {cout << char(i) << ":0" << endl;}else {int tmp = g[i][1] - g[i][0];for (int j = 2; j < ll; ++j) {tmp = min(tmp, g[i][j] - g[i][j-1]);}cout << char(i) << ":" << len - tmp << endl;}}for (int i = 'A'; i <= 'Z'; ++i) {if (g[i].empty()) continue;int ll = g[i].size();if (ll == 1) {cout << char(i) << ":0" << endl;}else {int tmp = g[i][1] - g[i][0];for (int j = 2; j < ll; ++j) {tmp = min(tmp, g[i][j] - g[i][j-1]);}cout << char(i) << ":" << len - tmp << endl;}}return 0;
}

F. Mo的极限

字符串处理统计次方和对应的系数，最后找最大的判断

分子上的系数可能全部为0
可能存在相同指数的多项式，需要去重
最简分数的分母是负数
最简分数的分母是1

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 1001;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;
const int mod = 1e9 + 7;
struct ac{int d, x;bool operator < (const ac &t) {return x > t.x;}
}a[maxn], b[maxn];void solve(string s, int &d, int &x) {int flag = 1;int len = s.size();if (s[0] == '-') flag = -1;int now = 0;if (s[0] == '+' || s[0] == '-') now = 1;int xx;for (int i = 0; i < len; ++i) {if (s[i] == '^') {xx = i + 1;break;}}int num = 0;for (int i = now; s[i] != 'x'; ++i) {num = num * 10 + s[i] - '0';}d = num * flag;num = 0;for (int i = xx; i < len; ++i) {num = num * 10 + s[i] - '0';}x = num;
}int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);  string s, t, tmp;cin >> s >> t;int lens, lent;int ls = s.size();int lt = t.size();int now = 1;map<int,int> mp;//处理第一个多项式for (int i = 0; i < ls; ++i) {if (i == ls-1 || (i && (s[i] == '+' || s[i] == '-'))) {if (i == ls-1) tmp += s[i];int d, x;solve(tmp, d, x);if (mp[x]) {//去重a[mp[x]].d += d;tmp = s[i];continue;}mp[x] = now;a[now].d = d;a[now++].x = x;tmp = s[i];}else tmp += s[i];}mp.clear();lens = now;now = 1;tmp = "";// 处理第二个多项式for (int i = 0; i < lt; ++i) {if (i == lt-1 || (i && (t[i] == '+' || t[i] == '-'))) {if (i == lt-1) tmp += t[i];int d, x;solve(tmp, d, x);if (mp[x]) {b[mp[x]].d += d;tmp = t[i];continue;}mp[x]=now;b[now].d = d;b[now++].x = x;tmp = t[i];}else tmp += t[i];}lent = now;sort(a+1, a + lens+1);sort(b+1, b + lent+1);int n, m = -1, a0, b0;for (int i = 1; i < lens; ++i) {if (a[i].d != 0) {a0 = a[i].d;            m = a[i].x;break;}}for (int i = 1; i < lent; ++i) {if (b[i].d != 0) {b0 = b[i].d;            n = b[i].x;break;}}    if (n > m) cout << 0 << endl;else if (n < m) cout << "oo\n";else {int tt = __gcd(abs(a0), abs(b0));a0 /= tt;b0 /= tt;if (b0 < 0) {b0 *= -1;a0 *= -1;}if (b0 == 1) cout << a0 << endl;else cout << a0 << "/" << b0 << endl;}return 0;
}

G. Mo的数学

容斥，初始化ans=n!ans = n!ans=n!，然后筛m的素因子，1−n1-n1−n中与mmm互质的数一定不含有mmm的素因子，容器奇数个素因子除掉，偶数个素因子乘上，每个素因子最多有nt\frac{n}{t}tn个
ans1t×2t×...×ntt=ansnt!tnt\frac{ans}{1t × 2t ×...×\frac{n}{t}t} = \frac{ans}{\frac{n}{t}!t^\frac{n}{t}}1t×2t×...×tntans=tn!ttnans
ans×1t×2t×...×ntt=ans×nt!tntans ×{1t × 2t ×...×\frac{n}{t}t} = ans ×{\frac{n}{t}!t^\frac{n}{t}}ans×1t×2t×...×tnt=ans×tn!ttn

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 1000001;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;
LL mod = 1e9 + 7;
LL jie[maxn];
LL exgcd(LL a, LL b, LL &x, LL &y) {LL d = a;if (b == 0) x = 1,y = 0;else {d = exgcd(b, a%b, y, x);y -= a / b * x;}return d;
}LL inv(LL a, LL mod) {LL x, y;LL d = exgcd(a, mod, x, y);return (d == 1) ? ((x + mod) % mod) : -1;
}LL pow(LL x, LL a) {LL ans = 1;while (a) {if (a&1) ans = ans * x % mod;x = x * x % mod;a >>= 1;}return ans;
}int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);  int n, m;jie[1] = jie[0] = 1;for (LL i = 2; i < maxn; ++i) {jie[i] = jie[i-1] * i % mod;}while (cin >> n >> m) {vector<LL> su;for (LL i = 2; i * i <= m; ++i) {if (m % i == 0) {su.push_back(i);while (m % i == 0) m /= i;}}if (m > 1) su.push_back(m);LL ans = jie[n];int len = su.size();for (int i = 1; i < (1<<len); ++i) {int cnt = 0;LL t = 1; for (int j = 0; (1<<j) <= i; ++j) {if (i & (1<<j)) {cnt++;t = t * su[j] % mod;}}LL x = jie[n/t] * pow(t, n/t) % mod;if (cnt & 1) ans = ans * inv(x, mod) % mod;else ans = ans * x % mod;}cout << ans << endl;        }return 0;
}

H. Mo的面积

只有两个矩形的矩形面积并，分类讨论相交矩形的长和宽

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
struct ac{int a, b, c, d;
};int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);  ac x, y;cin >> x.a >> x.b >> x.c >> x.d;cin >> y.a >> y.b >> y.c >> y.d;if (x.a > y.a) swap(x, y);int ans = 0;ans = (x.c - x.a) * (x.d - x.b);ans += (y.c - y.a) * (y.d - y.b);int l;if (y.a >= x.a && y.c <= x.c) l = y.c - y.a;if (y.a > x.a && y.c > x.c && y.a < x.c) l = x.c - y.a;if (y.a < x.a && y.c > x.a && y.c < x.c) l = y.c - x.a;if (y.a <= x.a && y.c >= x.c) l = x.c - x.a;if (y.c <= x.a || y.a >= x.c) l = 0;int r;if (y.b >= x.d || y.d <= x.b) r = 0;if (y.d >= x.d && y.b <= x.b) r = x.d - x.b;if (y.d <= x.d && y.b >= x.b) r = y.d - y.b;if (y.d > x.b && y.b < x.b && y.d < x.d) r = y.d - x.b;if (y.b > x.b && y.d > x.d && y.b < x.d) r = x.d - y.b;ans -= l * r;cout << ans << endl;return 0;
}

J. 简单递归

f[n]=2f[n−1]+f[n−2]2f[n] = 2f[n-1]+\frac{f[n-2]}{2}f[n]=2f[n−1]+2f[n−2]

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 105;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;
int mod = 1e9 + 7;
LL f[1000006];
int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);   int T;cin >> T;f[1] = f[2] = 1;for (int i = 3; i <= 1000000; ++i) {f[i] = (f[i-1] * 2 % mod) + (f[i-2] / 2 % mod);f[i] %= mod;}while (T--) {int n;cin >> n;cout << f[n] << endl;}return 0;
}

K. 高度期望

贪心把最小的树变成最大

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 105;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;int a[100005];
int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);   int n, m;cin >> n >> m;int sum = 0;for (int i = 0; i < n; ++i) {cin >> a[i];sum += a[i];}sort(a, a + n);int last = m * n - sum;int ans = 0;for (int i = 0; i < n && last > 0; ++i) {ans++;last -= 1000 - a[i];}cout << ans << endl;return 0;
}

L. 最优规划

并查集先把存在的边搞到一次，剩下的边从最小的边权开始看是否要加上

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#include <time.h>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i <= n; ++i)
const int maxn = 100001;
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std;int fa[maxn];
int find(int x) {return (x == fa[x]) ? x : fa[x] = find(fa[x]);
}int join(int x, int y) {int fx = find(x);int fy = find(y);if (fx == fy) return 0;if (fx > fy) swap(fx, fy);fa[fy] = fx;return 1;
}struct ac{LL u, v, d;bool operator < (const ac &t) {return d < t.d;}
}a[maxn];
int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);  int n, m, s;cin >> n >> m >> s;for (int i = 1; i <= n; ++i) fa[i] = i;for (int i = 0, u, v, d; i < m; ++i) {cin >> u >> v >> d;join(u, v);}LL ans = 0;for (int i = 0; i < s; ++i) {cin >> a[i].u >> a[i].v >> a[i].d;}sort(a, a + s);for (int i = 0; i < s; ++i) {int u = a[i].u;int v = a[i].v;LL d = a[i].d;int t = join(u, v);if (t) ans += d;}int flag = 1;for (int i = 1; i <= n && flag; ++i) {if (find(i) != find(1)) flag = 0;}if (flag) cout << ans << endl;else cout << "Concubines can't do it.\n";return 0;
}