LA 2678 Subsequence
利用单调性求最短连续子序列的和大于等于s；
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define maxn 100009 5 using namespace std; 6 7 int sum[maxn]; 8 int main() 9 { 10 int n,s,x; 11 while(scanf("%d%d",&n,&s)!=EOF) 12 { 13 memset(sum,0,sizeof sum); 14 int ans=n+1; 15 for(int i=1;i<=n;i++) 16 { 17 scanf("%d",&x); 18 sum[i]=sum[i-1]+x; 19 } 20 int j=1; 21 for(int i=1;i<=n;i++) 22 { 23 if(sum[j-1]>sum[i]-s)continue; 24 while(sum[j]<=sum[i]-s)j++; 25 ans=min(ans,i-j+1); 26 } 27 printf("%d\n",ans==n+1?0:ans); 28 } 29 return 0; 30 }
