51nod 1065 最小正子段和

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int N = 50001;
 6 const int inf  = 0x3f3f3f3f;
 7 pair<long long, int> sum[N];
 8 int a[N];
 9 int n;
10 int main(){
11     int i, j;
12     long long ans = inf;
13     scanf("%d", &n);
14     sum[0] = make_pair(0,0);
15     for(i = 1; i <= n; ++i){
16         scanf("%d", &a[i]);
17         sum[i].first = a[i] + sum[i-1].first;
18         sum[i].second = i;
19     }
20     sort(sum+1, sum+1+n);
21     for(i = 1; i <= n; ++i){
22         if(sum[i].first > 0)
23             ans = min(ans, sum[i].first);
24         if(sum[i].second > sum[i-1].second && sum[i].first > sum[i-1].first)
25             ans = min(ans, sum[i].first - sum[i-1].first);
26     }
27     printf("%lld\n", ans);
28     return 0;
29 }