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每日一题-Subsequence(前缀和+尺取法)(复习)
阅读量:319 次
发布时间:2019-03-04

本文共 521 字,大约阅读时间需要 1 分钟。

在这里插入图片描述

大意是:
给一个整数数列,找这个数列的子序列的和大于给定数值s的最短序列长度。

思路:

尺取法应用:
1.把子序列看成一个蚯蚓,当子序列和小于s时,头向前伸,直到子序列和大于等于s时就停止前伸,并记录长度。

2.尾部前伸,直到序列和小于数值s停止前伸,并记录长度。

然后再伸头,一直循环下去。

3.每次记录长度时更新结果,取最小值。

#include
using namespace std;typedef long long LL;int num[100005];void solve(){ int n,s; cin>>n>>s; for(int i=1;i<=n;i++) { cin>>num[i]; } LL sum=0; int len=n+1; for(int l=1,r=1;r<=n+1;) { if(sum>=s) { len = min(r-l,len); sum -= num[l]; l++;// if(l>=r) break; } if(sum
>t; while(t--) { solve(); } return 0; }

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