Educational Codeforces Round 24

A

好傻的，我还用了二分

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 2e5 + 10;LL n, k;bool Check(LL x){if(x == 0)return true;if((n / 2 - x) / x >= k)return true;return false;}int main(){cin >> n >> k;LL l = 0, r = n / 2, mid;LL ans = 0;while(l <= r){mid = (l + r) / 2;if(Check(mid)){ans = mid;l = mid + 1;}else{r = mid - 1;}}printf("%lld %lld %lld\n", ans, ans * k, n - (ans + ans * k));return 0;}

B

题意：n个数(1~n)组成的序列，m次操作，每次操作数i，然后到数字为i + ai的地方，如此操作下去，求符合要求的数组a

思路：直接枚举第一个数再数组a中的位置，然后check即可

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 2e5 + 10;int n, m;int num[105];int loca[105];int vis[105];map<int, bool> mp;int x[105];bool Check(){mst(vis, 0);for(int i = 1; i <= n; ++i){if(!num[i])continue;if(!vis[num[i]])vis[num[i]] = 1;elsereturn false;}return true;}int main(){cin >> n >> m;REP(i, 1, m){scanf("%d", loca + i);}bool f = false;for(int i = 1; i <= n; ++i){mst(num, 0);num[i] = loca[1];bool t = true;for(int j = 2; j <= m; ++j){int a = loca[j - 1], b = loca[j];int cnt = 0;while(1){++a, ++cnt;if(a > n)a = 1;if(a == b)break;}if(!num[loca[j - 1]]){num[loca[j - 1]] = cnt;}else{if(num[loca[j - 1]] != cnt){t = false;break;}}}/*for(int j = 1; j <= n; ++j)printf("%d ", num[j]);puts("");*/if(!t)continue;if(!Check())continue;if(t){f = true;break;}}if(!f){puts("-1");return 0;}for(int i = 1; i <= n; ++i){mp[num[i]] = true;}int cnt = 0;for(int i = 1; i <= n; ++i){if(!mp[i])x[cnt++] = i;}cnt = 0;for(int i = 1; i <= n; ++i){if(!num[i])num[i] = x[cnt++];}for(int i = 1; i <= n; ++i)printf("%d ", num[i]);puts("");return 0;}

C

题意：由d个沙发，分布在n × m的矩阵内，然后给出numl， numr， numt， numb分别代表某个沙发左边，右边，上边，下边的沙发数量，让你求出这个沙发的index，一个沙发占两个有公共边的格子。

思路：分四个方向分别统计前缀和后缀，然后枚举每一个沙发，看是否满足所给定的沙发数量

#include <cstdio>#include <cstring>#include <map>#include <algorithm>#include <iostream>#include <vector>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 2e5 + 10;int dirc[qq][4];int point[qq][4];int main(){int d, n, m;scanf("%d%d%d", &d, &n, &m);REP(i, 1, d){scanf("%d%d%d%d", &point[i][0], &point[i][2], &point[i][1], &point[i][3]);if(point[i][0] > point[i][1])swap(point[i][0], point[i][1]);if(point[i][2] > point[i][3])swap(point[i][2], point[i][3]);REP(j, 0, 3){dirc[point[i][j]][j]++;}}int cntl, cntr, cntt, cntb;scanf("%d%d%d%d", &cntl, &cntr, &cntt, &cntb);for(int i = 1; i <= n; ++i)dirc[i][0] += dirc[i - 1][0];for(int i = n; i >= 1; --i)dirc[i][1] += dirc[i + 1][1];for(int i = 1; i <= m; ++i)dirc[i][2] += dirc[i - 1][2];for(int i = m; i >= 1; --i)dirc[i][3] += dirc[i + 1][3];for(int i = 1; i <= d; ++i){int l = point[i][0], r = point[i][1], b = point[i][2], t = point[i][3];int numl = dirc[r - 1][0], numr = dirc[l + 1][1], numt = dirc[t - 1][2], numb = dirc[b + 1][3];if(l == r){numt--, numb--;}else{numl--, numr--;}if(numl == cntl && numr == cntr && numt == cntt && numb == cntb){printf("%d\n", i);return 0;}}puts("-1");return 0;}

D

题意：给定n个数，然后一个数字B，问是否存在一个数A，对于数字A的前缀和在任意一个i都满足cntA(i) >= cntB(i)，求出这个数。

思路：恩～～～感觉想到了就是很简单的，看代码吧。

我感觉实际上可以这样想，就是对于每一个数字B与数字B之间都要存在某个数，注意第一个B前面也要存在。

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int qq = 1e6 + 10;int num[qq];int n, A;map<int, int> mp;int main(){scanf("%d%d", &n, &A);for(int i = 1; i <= n; ++i){int x;scanf("%d", &x);if(x == A)num[A]++;else if(num[x] >= num[A])num[x]++;}for(int i = 1; i <= 1e6; ++i)if(i != A && num[i] >= num[A]){printf("%d\n", i);return 0;}puts("-1");return 0;}