POJ 3061 Subsequence（Two Pointers）
forever97
posted @ 2016年11月02日 00:26
in 算法-Two pointers
with tags
Two pointers
, 626 阅读
【题目链接】 http://poj.org/problem?id=3061
【题目大意】
给出S和一个长度为n的数列，问最短大于等于S的子区间的长度。
【题解】
利用双指针获取每一个恰好大于等于S的子区间，更新答案即可。
【代码】
#include <cstdio>
int T,a[100005];
int main(){
scanf("%d",&T);
while(T--){
int n,S,s,h,t,ans;
scanf("%d%d",&n,&S);ans=n;
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
for(int i=1,j=1;i<=n;i++){
s+=a[i];
if(s>=S){
ans=min(ans,i-j);
s-=a[j++];
}
}printf("%d\n",ans);
}return 0;
}
2024年1月18日 16:40
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