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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9422 | Accepted: 3797 |
Description
Input
Output
Sample Input
2 10 15 5 1 3 5 10 7 4 9 2 8 5 11 1 2 3 4 5
Sample Output
2 3
Source
ac代码
#include<stdio.h>
#include<string.h>
#define min(a,b) (a>b?b:a)
__int64 a[101000];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,i,j;
__int64 k;
scanf("%d%I64d",&n,&k);
a[0]=0;
for(i=1;i<=n;i++)
{
__int64 x;
scanf("%I64d",&x);
a[i]=a[i-1]+x;
}
int ans=100010;
// int i,j;
i=j=1;
while(i<=n&&j<=n)
{
while(j<=n&&a[j]-a[i-1]<k)
{
j++;
}
if(j<=n)
{
ans=min(ans,j-i+1);
i++;
}
else
break;
}
if(ans==100010)
ans=0;
printf("%d\n",ans);
}
}标签:
原文地址:http://blog.csdn.net/yu_ch_sh/article/details/45542733