POJ 3061 - Subsequence(尺取法)

最新推荐文章于 2020-04-08 20:36:23 发布

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本文介绍了一个经典的子序列和问题，即寻找连续子序列使得其和大于等于给定值的同时，区间长度最小。通过尺取法解决该问题，并提供了一段AC代码作为实现案例。

Subsequence
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12967 Accepted: 5479
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

题意：
给出n个数，要求求出这n个数中连续子段和大于等于m的情况下，区间的最小长度。

解题思路：
尺取法裸题，从头开始，如果符合条件，就去掉最左边的，否则加上后面的一个。

AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
const int maxn = 1e5+5;
int a[maxn];
void fun(int n,int m)
{
    int l = 1,r = 1,res = n+1,sum = 0;
    while(1)
    {
        while(sum < m && r <= n)    sum += a[r++];
        if(sum < m)   break;
        res = min(r-l,res);
        sum -= a[l++];
    }
    if(res == n+1)  printf("0\n");
    else            printf("%d\n",res);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i = 1;i <= n;i++)   scanf("%d",&a[i]);
        fun(n,m);
    }
    return 0;
}