Non-negative Partial Sums

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

3
2 2 1
3
-1 1 1
1
-1
0

3
2
0

#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000010;
int a[maxn*2];
int q[maxn*2];
int main()
{
    int n;
    while(scanf("%d", &n) && n)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            a[i+n] = a[i];
        }
        int cnt = 0;
        for(int i = 1; i <= 2*n; i++)
            a[i] += a[i-1];
        int front = 0, rear = -1;
        
        for(int i = 1; i < 2*n; i++)
        {
            while(front <= rear && i - q[front] >= n)
                front++;   
            while(front <= rear && a[i] <= a[q[rear]])
                rear--;
            q[++rear] = i;
            if(i >= n && a[q[front]] - a[i-n] >= 0)
                cnt++;
        }
        printf("%d\n", cnt);
    }
    return 0;
}

HDU 1193 Non-negative Partial Sums / 单调队列

原文：http://blog.csdn.net/u011686226/article/details/18984465