二分法中的两个函数-【lower_bound】【upper_bound】

最新推荐文章于 2024-10-06 22:43:48 发布

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本文介绍C++ STL中lower_bound和upper_bound函数的应用，并通过一个具体示例展示如何利用这些函数解决实际问题。示例为寻找序列中元素之和大于等于给定值的最短连续子序列。

/*
lower_bound(begin(),end()+1,k);
upper_bound(begin(),end()+1,k);
*/
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
	int num[20] = {1,2,3,4,5,5,5,6,7,10};
//	int k;
//	scanf ("%d",&k);
	int k = 20;
	int pos1 = lower_bound(num,num+10,k) - num;   //找第一个数出现的坐标   第一个5 输出为4 
	printf ("%d\n",pos1);
	int pos2 = upper_bound(num,num+10,k) - num;   //找最后一个数坐标   需在num后减 1 
	printf ("%d\n",pos2-pos1);                   //为这个数出现的个数 
	return 0;
}

例题

【POJ】2061 - Subsequence

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

大佬代码

#include <cstdio>  
#include <algorithm>  
#define MAX 100000  
using namespace std;  
int main()  
{  
    int u;  
    int sum[MAX+22];  
    int ans;  
    int n,S;  
    scanf ("%d",&u);  
    while (u--)  
    {  
        scanf ("%d %d",&n,&S);  
        scanf ("%d",&sum[0]);  
        for (int i = 1 ; i < n ; i++)  
        {  
            scanf ("%d",&sum[i]);  
            sum[i] += sum[i-1];  
        }  
        if (sum[n-1] < S)  
        {  
            printf ("0\n");  
            continue;  
        }  
        ans = n;  
        int pos;        //存放序列开始的位置   
        for (int i = n - 1 ; i >= 0 ; i--)       //从最后面开始，把其当作序列尾   
        {  
            if (sum[i] < S)  
                break;  
            pos = upper_bound(sum , sum + n , sum[i] - S) - sum;  
            ans = min(ans , i - pos + 1);  
        }  
        printf ("%d\n",ans);  
    }  
    return 0;  
}