POJ 3061 Subsequence （尺取法）

最新推荐文章于 2020-05-30 15:12:50 发布

玻璃年华Alex

最新推荐文章于 2020-05-30 15:12:50 发布

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本文介绍了一种算法，用于寻找给定整数序列中最短的连续子序列，该子序列的元素之和大于或等于指定的目标值。文章提供了一个完整的C++实现案例，并通过样例输入输出展示了算法的应用。

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Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 7604Accepted: 2879

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

附：

#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <set>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define Max_N 100008
#define Max_S 100000008
 
using namespace std;
int g_N, g_S;
int g_Seq[Max_N];
 
int FindSubsequence(){
    int res = g_N + 1;
    int left = 1, right = 1, sum = 0;
    while(1){
        while(right <= g_N&&sum < g_S){
            sum += g_Seq[right++];
        }
        if(sum < g_S) break;
        res =  Min(res , right - left);
        sum -= g_Seq[left++];
    }
    return res > g_N?0:res;
}
int main()
{
    int test;
    cin>>test;
    while(test--){
        scanf("%d%d",&g_N,&g_S);
        for(int i = 1;i <= g_N;i++){
            scanf("%d",&g_Seq[i]);
        }
        printf("%d\n",FindSubsequence());
    }
    return 0;
}