POJ：3061 Subsequence（尺取法）

A - Subsequence

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.

Input

The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.

Output

For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

Sample Input

210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5

Sample Output

23

题意：给你一个数列和一个数字q，找出这个数列中一个连续的数列，使该数列的和>=q，求这个连续数列的最小长度。

用尺取法：

尺取法

反复地推进区间的开头和末尾，来求满足条件的最小区间的方法称为尺取法。

主要思想为：当a1,  a2  , a3 满足和>=S,得到一个区间长度3，那么去掉开头a1,   剩下 a2,a3，判断是否满足>=S,如果满足，那么区间长度更新，如果不满足，那么尾部向后拓展，判断a2,a3,a4是否满足条件。重复这样的操作。

个人对尺取法的理解：

当一个区间满足条件时，那么去掉区间开头第一个数，得到新区间，判断新区间是否满足条件，如果不满足条件，那么区间末尾向后扩展，直到满足条件为之，这样就得到了许多满足条件的区间，再根据题意要求什么，就可以在这些区间中进行选择，比如区间最长，区间最短什么的。这样跑一遍下来，时间复杂度为O（n)。

#include <stdio.h>#define INF 0x3f3f3f3fint a[100010];int min(int x,int y){if(x>y)return y;return x;}int main(){int t,n,q;scanf("%d",&t);while(t--){scanf("%d%d",&n,&q);for(int i=0;i<n;i++){scanf("%d",&a[i]);}int ans=INF;int l=0,r=0;//左右区间 int sum=0;while(1){while(sum<q&&r<n)//向右扩展，找出大于等于q的区间 {sum=sum+a[r];r++;}if(sum<q)//意思是扫完了，结果还是小于q,则说明不存在 break;while(l<=r&&sum>=q)//在所找出的区间中 ，依次去掉最左端，看能否满足sum大于等于q {sum=sum-a[l];l++;}ans=min(ans,r-l+1);}if(ans==INF){printf("0\n");}else{printf("%d\n",ans);}}return 0;}