hihoCoder 1051 : 补提交卡 枚举

              思路：预处理cnt(i)表示前i个数中有多少天需要补提交卡，枚举各个连续区间，区间[j, i]中需要补提交卡的天数是cnt(i) - cnt(j-1)，判断m是否大于等于cnt(i) - cnt(j-1)，更新答案即可。

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 100 + 5;int a[maxn], cnt[maxn];int main() {int T;scanf("%d", &T);while(T--) {memset(a, 0, sizeof(a));int n, m;scanf("%d%d", &n, &m);int x;for(int i = 1; i <= n; ++i) {scanf("%d", &x);a[x] = 1;}cnt[0] = 0;for(int i = 1; i <= 100; ++i) {cnt[i] = cnt[i-1];if(a[i]) cnt[i]++;}int ans = 0;for(int i = 1; i <= 100; ++i) {for(int j = 1; j <= i; ++j) {int tol = cnt[i] - cnt[j-1];if(m >= tol) ans = max(ans, i - j + 1);}} printf("%d\n", ans);}return 0;}

如有不当之处欢迎指出！