枚举一个点,假设它一定符合条件,据此珂以\(O(n)\)算出要删去几个点
于是就\(O(n^2)\)解决了,貌似加一个前缀和可以在\(O(n)\)的时间复杂度内解决本问题,但对于这个数据范围来说\(O(n^2)\)戳戳有余
放个代码:
#include <cstdio>
#define ll long long
inline ll read(){
ll x = 0; int zf = 1; char ch = ' ';
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}
int a[105];
int main(){
int n = read();
for (int i = 1; i <= n; ++i)
a[i] = read();
int ans = 0, cnt;
for (int i = 1; i <= n; ++i){
cnt = 0;
for (int j = i + 1; j <= n; ++j)
if (!a[j])
++cnt;
for (int j = 1; j < i; ++j)
if (a[j])
++cnt;
if (n - cnt > ans)
ans = n - cnt;
}
printf("%d", ans);
return 0;
}
原文地址:https://www.cnblogs.com/linzhengmin/p/10861319.html
时间: 2024-11-05 10:07:33