POJ 3061 Subsequence(Two Pointers)

forever97 posted @ 2016年11月02日 00:26 in 算法-Two pointers with tags Two pointers , 787 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;
}
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pavzi.com 说:
2024年1月18日 16:40

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