HDU

保证有解的情况下，对于每一个负数，前面都存在几个正数使之前缀和大于零

从后往前找一遍，遇到负数往前加

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <set>#include <queue>#include <algorithm>#include <vector>#include <cctype>#include <stack>#include <sstream>#include <list>#include <map>#include <assert.h>#define debug() puts("************")#define MS(a,b) memset(a,b,sizeof a)using namespace std;typedef long long ll;typedef pair<int,int> P;const int INF = 0x3f3f3f3f;const double inf = 1e20;const double eps = 1e-15;const int dx[] = {-1,0,1,0};const int dy[] = {0,1,0,-1};const int maxn = 1000000 + 7, maxd = 100000 + 7;int n;ll a[maxn];int main() {    while(scanf("%d", &n) != EOF && n) {        for(int i = 0; i < n; ++i)        scanf("%lld", &a[i]);        int ans = 0;        for(int i = n-1; i >= 0; --i) {            if(a[i] >= 0) ans++;            else a[i-1] += a[i];        }        printf("%d\n", ans);    }    return 0;}