fzu 1914 Funny Positive Sequence

题目链接：fzu 1914 Funny Positive Sequence

题目大意：给出n，以及长度为n的数列a[i]，现在以每个i为起点，计算d[i][j]表示以i为起点，计算后面j个数的和（如果i + j超过n，从1开始继续计算）问有几个i满足d[i][j]（1≤j≤n)均大于0。

解题思路：这题还是有点技巧的，从数组的后面枚举，碰到a[i] ≤ 0，就开始往前叠加，凡是和小于等于0的都是不满足情况的位置i。注意一个坑点就是，当叠加到i为0，即数组到头的时候sum仍小于0的时候，要考虑从后再遍历一下数组。

#include <stdio.h>#include <string.h>#define ll long longconst int N = 500005;int n, v[N];ll a[N];void init() {scanf("%d", &n);memset(v, 0, sizeof(v));for (int i = 0; i < n; i++)scanf("%lld", &a[i]);}int solve() {int ans = n;ll sum = 0;for (int i = n - 1; i >= 0 && ans; i--) {if (sum <= 0) {sum += a[i];if (sum <= 0) {ans--;v[i] = 1;}} else if (a[i] <= 0) {sum = a[i];ans--;v[i] = 1;}}int i = n - 1;while (sum <= 0) {sum += a[i];if (sum <= 0 && !v[i]) {v[i] = 1;ans--;}i--;if (ans == 0 || i < 0) break;}return ans;}int main () {int cas;scanf("%d", &cas);for (int i = 1; i <= cas; i++) {init();printf("Case %d: %d\n", i, solve());}return 0;}