POJ3061 Subsequence

Posted 2020-10-05 小时のblog

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16520		Accepted: 7008

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

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

Southeastern Europe 2006

 

题目大意：t个测试数据 对于一个长度为N的序列 求出总和不小于S的连续子序列的长度的最小值。

题解：

解法一、

二分搜索

前缀和是满足单调性的，计算从每一个数开始总和刚好大于s的长度。具体实现就是

二分搜索s[i]+s是否存在于前缀和数组中，就是查找以i+1开头的总和刚好大于s的最短长度。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int t,n,s,x;
int sum[100005];
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&s);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            sum[i]=sum[i-1]+x;
        }
        if(sum[n]<s){
            printf("0\n");
            continue;
        }
        int res=n;
        for(int i=0;sum[i]+s<=sum[n];i++){
            int t=lower_bound(sum+i+1,sum+n+1,sum[i]+s)-sum;
            res=min(res,t-i);
        }
        printf("%d\n",res);
    }
    return 0;
}

 

时间复杂度nlog(n)

解法二、

尺取法 时间复杂度O（n）

#include<iostream>
#include<cstdio>
using namespace std;
int res,t,n,s,l,r,sum;
int a[100006]; 
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&s);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        l=r=1;sum=0;res=n+1;
        for(;;){
            while(r<=n&&sum<s){
                sum+=a[r++];
            }
            if(sum<s)break;//必须先判断才能更新解 
            res=min(res,r-l);
            sum-=a[l++];
        }
        if(res>n)printf("0\n");
        else printf("%d\n",res);
    }
    return 0;
}