A ⼩Hi的钟表

算秒数，算角度，注意180以内，所以大于180之后拿360减

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
int h, m, s, t;int main(){//freopen("in.txt", "r", stdin);int _;scanf("%d", &_);for (; _--;){scanf("%d%d%d%d", &h, &m, &s, &t);t += h * 3600 + m * 60 + s;t %= 3600 * 12;m = t % 3600;double ans = fabs(1.0 * t / 120 - 1.0 * m / 10); if (ans > 180) ans = 360 - ans; //printf("t = %d\n", t);//double ans = 1.0 * t / 120;printf("%.4lf\n", ans);}return 0;
}

B 影院座位

没想到这题居然是最难的一题，卡了一万年

hint: 中位数，前缀和

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 1e6 + 5;
int a[MAXN];
int Getc[MAXN];
int Sum[MAXN];
int main(){//freopen("in.txt", "r", stdin);int N, M;cin >> N >> M;for(int i = 1; i <= N; i++) {scanf("%d", Getc + i);}int cnt = 0;for(int i = 1; i <= N; i++) {if(Getc[i] == 0) {a[cnt++] = i;}}for(int i = 0; i < cnt; i++) {a[cnt + i] = a[i] + N;}Sum[0] = a[0];for(int i = 1; i < cnt * 2; i++) {Sum[i] = Sum[i - 1] + a[i];}if(cnt < M) {printf("-1\n");return 0;}int ans = a[M / 2] * (M / 2) - Sum[M / 2 - 1] + Sum[M - 1] - Sum[M / 2] - (M - 1 - M / 2) * a[M / 2];for(int i = 1; i < cnt; i++) {ans = min(ans, a[(i + i + M) / 2] * (M / 2) - Sum[(i + i + M) / 2 - 1] + Sum[i - 1] + Sum[M + i - 1] - Sum[(i + i + M) / 2] - (M - 1 - M / 2) * a[(i + i + M) / 2]);}printf("%d\n", ans);return 0;
}

C 同步H公司的终端

树形DP，统计每个点到叶子的最大距离，让每个子树都是一个“相等”、“平衡”状态

尽量修改靠上的节点，这样最后修改结果最小

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e5 + 7;
LL a[N], ans;
int n;
bool NotRoot[N];
int FindRoot(){for (int i = 1; i <= n; i++) if (!NotRoot[i]) return i;return -1; 
} struct Edge{int from, to, nxt;Edge(){}Edge(int u, int v, int n):from(u), to(v), nxt(n){}
} edges[N * 2];
int E, head[N]; 
LL dist[N];
inline void AddEdge(int f, int t){edges[++E] = Edge(f, t, head[f]);head[f] = E;
}
inline void Init(){E = -1;for (int i = 0; i <= n; i++) head[i] = -1;for (int i = 0; i <= n; i++) dist[i] = 0;
}int dfs(int u, int pre){//printf("dfs(%d, %d)\n", u, pre);LL Max = 0;for (int i = head[u]; i != -1; i = edges[i].nxt){Edge &e = edges[i];dfs(e.to, u);Max = max(Max, dist[e.to]);}dist[u] = Max + a[u];for (int i = head[u]; i != -1; i = edges[i].nxt){Edge &e = edges[i];ans += Max - dist[e.to];}
}int main(){//freopen("in.txt", "r", stdin);int u, v;for (; ~scanf("%d", &n);){a[0] = 0;for (int i = 1; i <= n; i++) scanf("%d", &a[i]);Init();for (int i = 1; i < n; i++) {scanf("%d%d", &u, &v);AddEdge(u, v);NotRoot[v] = 1;}int root = FindRoot();ans = 0;dfs(root, -1);printf("%lld\n", ans);}return 0;
}

D 大富翁

本以为是个压轴题，以为要线段树什么各种数据结构来维护

然后一看过的人多的不正常，想，先拿30分再说，然后就莫名其妙过了

标算应该不是这么粗暴的，加强数据应该还是过不了

这里就是根据题意模拟

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 1e6 + 5;
int a[MAXN];
int N;
int Find(int x) {if(x > N) return 0;return 1 + Find(x + a[x]);
}
int main(){cin >> N;for(int i = 1; i <= N; i++) {scanf("%d", a + i);}int Q;cin >> Q;while(Q--) {int x;scanf("%d", &x);if(x == 1) {int y;scanf("%d", &y);printf("%d\n", Find(y));} else {int u, v;scanf("%d%d", &u, &v);a[u] = v;}}return 0;
}

hiho一下第165周

分隔相同字符

复杂度1e5*26*26

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 7;
char st[N];
int Hash[222], n;int getMax(char ch){int ans = 0;for (char i = 'a'; i <= 'z'; i++){if (i == ch) ans = max(ans, Hash[i]-1);else ans = max(ans, Hash[i]);}return ans;
}char FindChar(char last){n--;for (char i = 'a'; i <= 'z'; i++){if (!Hash[i]) continue;if (i == last) continue;int Max = getMax(i);if (Max * 2 > n + 1) continue;Hash[i]--;return i;}return '0';
}int main(){//freopen("in.txt", "r", stdin);scanf("%s", st);int len = strlen(st);n = len;memset(Hash, 0, sizeof(Hash));for (int i = 0; i < len; i++) Hash[st[i]]++;int Max = getMax('0');if (Max * 2 > len + 1) puts("INVALID");else {char last = ' ';for (int i = 0; i < len; i++){last = FindChar(last);printf("%c", last);}puts("");}return 0;
}