Codeforces Round #419 (Div. 2)

A

模拟一下即可、

#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 = 1e5 + 10;int main(){int a, b;scanf("%d:%d", &a, &b);int cnt = 0;while(true){int t2 = b % 10;int t1 = b / 10;int ans = t1 + t2 * 10;if(a == ans){printf("%d\n", cnt);break;}cnt++;b++;if(b >= 60){b = 0;a++;}if(a >= 24){a = 0;}}return 0;}

B

题意：给出n个区间和q次查询和一个数k，给出n个区间ai，bi表示区间ai-bi之间的数都加一，q次查询区间ai-bi有多少个数超过k。

思路：参考传送门

#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, k, q;int num[qq];int sum[qq];int main(){scanf("%d%d%d", &n, &k, &q);int a, b;for(int i = 1; i <= n; ++i){scanf("%d%d", &a, &b);num[a] += 1, num[b + 1] += -1;}for(int i = 1; i <= 200000; ++i)sum[i] = sum[i - 1] + num[i];for(int i = 1; i <= 200000; ++i){if(sum[i] < k)sum[i] = 0;elsesum[i] = 1;}for(int i = 1; i <= 200000; ++i)sum[i] += sum[i - 1];while(q--){scanf("%d%d", &a, &b);printf("%d\n", max(sum[b] - sum[a - 1], 0));}return 0;}

C

题意：给出n，m的矩阵，每次可以选择一行或者一列进行加一，要求你在最操作数最少的情况下得到所给矩阵，没办法得出矩阵则输出-1

思路：其实挺简单的，最开始对这个操作数最少不理解，但想了想这个样例以后

2 3

1 1 1

1 1 1

会发现此时满足行就是两次，满足列就是三次，所以首先判断一下行列的大小决定先满足谁即可

#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 = 100 + 10;int n, m;int num[qq][qq];vector<int> ans[2];void Solve(int x, int y, int z){REP(i, 1, x){int minx = 1e9;REP(j, 1, y)minx = min(minx, z ? num[i][j] : num[j][i]);REP(j, 1, y)(z ? num[i][j] : num[j][i]) -= minx;while(minx > 0)ans[z].pb(i), minx--;}}int main(){scanf("%d%d", &n, &m);REP(i, 1, n)REP(j, 1, m)scanf("%d", &num[i][j]);if(n <= m)Solve(n, m, 1), Solve(m, n, 0);elseSolve(m, n, 0), Solve(n, m, 1);REP(i, 1, n)REP(j, 1, m)if(num[i][j]){printf("-1\n");return 0;}printf("%d\n", ans[0].size() + ans[1].size());REP(i, 1, ans[0].size())printf("col %d\n", ans[0][i - 1]);REP(i, 1, ans[1].size())printf("row %d\n", ans[1][i - 1]);return 0;}