HDU2058

最新推荐文章于 2019-03-26 19:47:42 发布

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http://acm.hdu.edu.cn/showproblem.php?pid=2058

The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10174 Accepted Submission(s): 3111

Problem Description

Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.

Input

Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.

Output

For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.

Sample Input

20 10
50 30
0 0

Sample Output

[1,4]
[10,10]

[4,8]
[6,9]
[9,11]
[30,30]

找一个子序列的和等于M，那么这个子序列可以看成a+1, a+2, ... , a+d 这时，d就为这个序列的长度,起始数字就是a+1，而这个序列的和即M=a*d + (1 + d) * d /2;得出d*d<2 * m，从而可以枚举d,计算出a

#include<iostream>
#include<cmath>

using namespace std;

int main(){
        int n,m;
        int d,b;
        while(cin>>n>>m && (n||m)){
                for(d=sqrt(2.0 *m); d>0;d--){
                        b=m-(d+d*d)/2;
                        if(b%d==0)
                        cout<<"["<<(b/d)+1<<","<<(b/d)+d<<"]"<<endl;
                }
                cout<<endl;
        }
        return 0;
}

网络上的其他相关代码：

#include <stdio.h>
#include <math.h>
int main()
{
	int n, m, len, a;
	while(~scanf("%d%d", &n,&m))
	{
		if(!n && !m)
			break;
		len = sqrt(double(m * 2)) + 1;
		//根据等差数列来算,数列长度
		// m = (a + a + len - 1) * len / 2
		// m = a * len + len(len - 1)/ 2
		// m - len * (len - 1) / 2 = a * len
		//a = m / len - (len - 1) / 2
		while(--len)
		{
			a = m / len - (len - 1) / 2;
			if((a + a + len - 1) * len / 2 == m)
			{
				printf("[%d,%d]\n", a, a+len - 1);
			}
		}
		printf("\n");
	}
	return 0;
}