题解报告（CDUT暑期集训——第一场）

A - Maximum Multiple

HDU - 6298

• 思路：先按照题意打表 发现规律 就出来了（最开始没开long long贡献了3发 然后又忘了换行又贡献了一发

• AC代码

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;ll Pow(ll a, ll b, ll c){ll ans = 1;a %= c;while (b){if (b & 1) ans = (ans * a) % c;a = (a * a) % c;b >>= 1;}return (ans % c);
}bool cmp(const int &a, const int &b){return a < b;
}ll gcd(ll x, ll y){return x % y == 0 ? y : gcd(y, x % y);
}ll lcm(ll x, ll y){return x / gcd(x, y) * y;
}const int MAXN = 2000 + 5;
const int mod = 1e9 + 7;
ll comb[MAXN][MAXN];void init(){for (int i = 0; i < MAXN; i ++ ){comb[i][0] = comb[i][i] = 1;for (int j = 1; j < i; j ++ ){comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];comb[i][j] %= mod;}}
}ll mult_mod(ll a, ll b, ll c){a %= c;b %= c;ll res = 0;while (b){if (b & 1){res += a;res %= c;}a <<= 1;a %= c;b >>= 1;}return res;
}int main(){ll t;scanf("%lld", &t);ll n;while (t--){scanf("%lld", &n);if (n % 12 == 1 || n % 12 == 2 || n % 12 == 5 || n % 12 == 7 || n % 12 == 10 || n % 12 == 11) printf("-1\n");else{if (n % 3) printf("%lld", (n / 4) * (n / 4) * (n / 4) * 2);else printf("%lld", (n / 3) * (n / 3) * (n / 3));}}return 0;
}

B - Balanced Sequence

HDU - 6299

• 思路：贪心 先把每个串里面的括号匹配之后去掉 再对它们排序 匹配剩下的括号就行了（开始莫名其妙MLE了九发 找不到原因 然后再交TLE RE 后面发现cmp写错了 改了之后就过了

• AC代码

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string>
#include<string.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;const int N = 100010;int n, t, ans, cnt_l;
char c[N];struct st{int l;int r;
}s[N];bool cmp(st a, st b){if (a.l > a.r && b.l <= b.r) return true;else if (a.l <= a.r && b.l > b.r) return false;else if (a.l > a.r && b.l > b.r) return a.r < b.r;else return a.l > b.l;
}int main(){scanf("%d", &t);while (t --){ans = 0;cnt_l = 0;scanf("%d", &n);for (int i = 1; i <= n; i ++ ){s[i].l = 0;s[i].r = 0;scanf("%s", c);for (int j = 0; j < strlen(c); j ++ ){if (c[j] == '(') s[i].l ++;if (c[j] == ')') s[i].l == 0 ? s[i].r ++ : (s[i].l -= 1, ans += 2);}}sort(s + 1, s + n + 1, cmp);for (int i = 1; i <= n; i ++ ){if (s[i].r > cnt_l) s[i].r = cnt_l;ans += 2 * s[i].r;cnt_l += s[i].l - s[i].r;}printf("%d\n", ans);}return 0;
}

C - Triangle Partition

HDU - 6300

• 思路：贪心 写个struct对x排个序就出来了（

• AC代码

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;ll Pow(ll a, ll b, ll c){ll ans = 1;a %= c;while (b){if (b & 1) ans = (ans * a) % c;a = (a * a) % c;b >>= 1;}return (ans % c);
}//bool cmp(const int &a, const int &b){
//    return a < b;
//}ll gcd(ll x, ll y){return x % y == 0 ? y : gcd(y, x % y);
}ll lcm(ll x, ll y){return x / gcd(x, y) * y;
}const int MAXN = 2000 + 5;
const int mod = 1e9 + 7;
ll comb[MAXN][MAXN];void init(){for (int i = 0; i < MAXN; i ++ ){comb[i][0] = comb[i][i] = 1;for (int j = 1; j < i; j ++ ){comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];comb[i][j] %= mod;}}
}ll mult_mod(ll a, ll b, ll c){a %= c;b %= c;ll res = 0;while (b){if (b & 1){res += a;res %= c;}a <<= 1;a %= c;b >>= 1;}return res;
}const int N = 1010;struct point{int x, y;int index;
}p[N];bool cmp(point a, point b){if (a.x == b.x) return a.y < b.y;return a.x < b.x;
}int main(){int t;scanf("%d", &t);int n;while (t--){scanf("%d", &n);for (int i = 1; i <= 3 * n; i ++ ){scanf("%d%d", &p[i].x, &p[i].y);p[i].index = i;}sort(p + 1, p + 3 * n + 1, cmp);for (int i = 1; i <= n; i ++ )printf("%d %d %d\n", p[(i - 1) * 3 + 1].index, p[(i - 1) * 3 + 2].index, p[(i - 1) * 3 + 3].index);}return 0;
}

D - Distinct Values

HDU - 6301

• 思路：贪心 先排序 后用set维护就行了（学到了一个新东西 迭代器赋值前要加* 要不然会报错

• AC代码

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<set>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;ll Pow_mod(ll a, ll b, ll c){ll ans = 1;a %= c;while (b){if (b & 1) ans = (ans * a) % c;a = (a * a) % c;b >>= 1;}return (ans % c);
}const int N = 1000010;int t, n, m, L, R;
set<int> st;
int ans[N];struct node{int l, r;
}interval[N];int cmp(node a, node b){return a.l == b.l ? a.r < b.r : a.l < b.l;
}int main(){scanf("%d", &t);while (t -- ){st.clear();scanf("%d%d", &n, &m);for (int i = 1; i <= m; i ++ )scanf("%d%d", &interval[i].l, &interval[i].r);sort(interval + 1, interval + m + 1, cmp);L = interval[1].l;R = interval[1].r;for (int i = 1; i <= n; i ++ ){st.insert(i);ans[i] = 1;}for (int i = L; i <= R; i ++ ){ans[i] = *st.begin();st.erase(st.begin());}for (int i = 2; i <= m; i ++ ){while (L < interval[i].l){st.insert(ans[L]);L ++;}while (R < interval[i].r){if (R < interval[i].l - 1) R ++;else{R ++;ans[R] = *st.begin();st.erase(st.begin());}}}for (int i = 1; i < n; i ++ ) printf("%d ", ans[i]);printf("%d\n", ans[n]);}
}

G - Chiaki Sequence Revisited

HDU - 6304

• 思路：先打表 找规律 令f(i)为i出现的次数 再打表 去掉a1 = 1出现的一次 可以得到f(2 * i) = f(i) + 1 f(2 * i + 1) = 1 当时到了这里就卡死了 卡了两个半小时 然后想着把f(i) = n, n = 1, 2, 3, ...打印出来看下 结果一打印就发现了f(i)相同的构成了一个首项为2^(n-1) 公差为2^n的等差数列 就直接根据公式就解出来了 结束前没能写出来 结束之后几分钟写好交了三发才过 第一发用自己写的乘法取膜函数tle了 第二发有一处忘了取膜贡献了一发wa 第三发才终于过了

• AC代码

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;ll Pow_mod(ll a, ll b, ll c){ll ans = 1;a %= c;while (b){if (b & 1) ans = (ans * a) % c;a = (a * a) % c;b >>= 1;}return (ans % c);
}ll Pow(ll a, ll b){ll ans = 1;while (b){if (b & 1) ans *= a;a *= a;b >>= 1;}return ans;
}bool cmp(const int &a, const int &b){return a < b;
}ll gcd(ll x, ll y){return x % y == 0 ? y : gcd(y, x % y);
}ll lcm(ll x, ll y){return x / gcd(x, y) * y;
}const int MAXN = 2000 + 5;
const int mod = 1e9 + 7;
ll comb[MAXN][MAXN];void init(){for (int i = 0; i < MAXN; i ++ ){comb[i][0] = comb[i][i] = 1;for (int j = 1; j < i; j ++ ){comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];comb[i][j] %= mod;}}
}ll mult_mod(ll a, ll b, ll c){a %= c;b %= c;ll res = 0;while (b){if (b & 1){res += a;res %= c;}a <<= 1;a %= c;b >>= 1;}return res;
}const int N = 63;ll s[N], t[N];ll calc(ll n){ll tmp = 0;if (n == 1) return 1;n -= 1;for (int i = N - 1; i >= 0; i -- ){while (s[i] <= n){n -= s[i];tmp += t[i];}}return tmp;
}ll get_pos(ll n){ll pos = 0;if (n == 1) return 1;n -= 1;for (int i = N - 1; i >= 0; i -- ){while (t[i] <= n){n -= t[i];pos += s[i];}}return pos + 1;
}int main(){s[0] = 1;t[0] = 1;for (int i = 1; i < N; i++){s[i] = s[i - 1] * 2 + 1;t[i] = t[i - 1] * 2;}int T;scanf("%d", &T);ll n;while (T--){scanf("%lld", &n);ll tmp = calc(n);ll cnt = n - get_pos(tmp);ll sum = 0;for (int i = 1; t[i - 1] <= tmp; i ++ ){ll b1 = t[i - 1];ll d = t[i];ll n_ = (tmp - b1) / d;if ((tmp - b1) % d) n_ += 1;ll bn = b1 + (n_ - 1) * d;ll sum_b = (b1 % mod + bn % mod) * (n_ % mod) * 500000004 % mod;sum += i * sum_b % mod;sum %= mod;if (!((tmp - b1) % d)) sum += (cnt % mod) * (tmp % mod) % mod;sum %= mod;}printf("%lld\n", sum + 1);}return 0;
}

K - Time Zone

HDU - 6308

• 思路：数学题 把时区差换算成时间差（分钟） 然后换算过去就行了 最开始写了个贼蠢的模拟 写的很长 还wa了 后面才想起来没写%02d才wa的

• AC代码

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;ll Pow(ll a, ll b, ll c){ll ans = 1;a %= c;while (b){if (b & 1) ans = (ans * a) % c;a = (a * a) % c;b >>= 1;}return (ans % c);
}bool cmp(const int &a, const int &b){return a < b;
}ll gcd(ll x, ll y){return x % y == 0 ? y : gcd(y, x % y);
}ll lcm(ll x, ll y){return x / gcd(x, y) * y;
}const int MAXN = 2000 + 5;
const int mod = 1e9 + 7;
ll comb[MAXN][MAXN];void init(){for (int i = 0; i < MAXN; i ++ ){comb[i][0] = comb[i][i] = 1;for (int j = 1; j < i; j ++ ){comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];comb[i][j] %= mod;}}
}ll mult_mod(ll a, ll b, ll c){a %= c;b %= c;ll res = 0;while (b){if (b & 1){res += a;res %= c;}a <<= 1;a %= c;b >>= 1;}return res;
}int main(){int t;scanf("%d", &t);int a, b;while (t--){char flag;double utc;scanf("%d %d UTC%c%lf", &a, &b, &flag, &utc);int tmp = (int)(utc * 10 + 0.1);if (flag == '-') tmp = -tmp;int sum = a * 60 + b + (tmp * 6 - 8 * 60);sum %= (24 * 60);if (sum < 0) sum += (24 * 60);printf("%02d:%02d\n", sum / 60, sum % 60);}return 0;
}