1209. Sequence Sum Possibi

最新推荐文章于 2025-08-18 20:25:51 发布

原创最新推荐文章于 2025-08-18 20:25:51 发布 · 931 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#input #output #each #integer #less

刷题之旅同时被 2 个专栏收录

176 篇文章

订阅专栏

西西里水题（水到爆阿）

125 篇文章

订阅专栏

本文探讨了如何通过编程计算特定数值可以表示为连续正整数序列的多种方式。通过数学推导和算法实现，展示了求解此类问题的方法，并提供了样例输入输出以验证解决方案的有效性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1209. Sequence Sum Possibi

Description

Most positive integers may be written as a sum of a sequence of at least two consecutive positive integers. For instance,

6 = 1 + 2 + 3
9 = 5 + 4 = 2 + 3 + 4

but 8 cannot be so written.

Write a program which will compute how many different ways an input number may be written as a sum of a sequence of at least two consecutive positive integers.

Input

The first line of input will contain the number of problem instances N on a line by itself, (1<=N<=1000) . This will be followed by N lines, one for each problem instance. Each problem line will have the problem number, a single space and the number to be written as a sequence of consecutive positive integers. The second number will be less than 2^31 (so will fit in a 32-bit integer).

Output

The output for each problem instance will be a single line containing the problem number, a single space and the number of ways the input number can be written as a sequence of consecutive positive integers.

Sample Input

7 
1 6 
2 9 
3 8 
4 1800 
5 987654321 
6 987654323 
7 987654325

Sample Output

纯数学题。

设 n+.. +(n+k)=X， n>0,k>0

有 (2n+k)(k+1)=2X

k(k+1) <2X

k^2<2X

k<sqrt(2X)

枚举k到sqrt(2X)就行了

假设num = (a + 0) + (a + 1) + ... + (a + i - 1) 其中i个数，则
num = i * a + 1 + 2 + ... + i - 1 = a*i + (i - 1) * i / 2
故num - (i - 1) * i / 2 = a * i
a为整数，num - (i - 1) * i / 2可被i整除

#include<iostream>
using namespace std;
int main()
{
	int n;
	cin>>n;
	int case_index;
	int num;
	for(int i=1;i<=n;i++)
	{
		int count=0;
		cin>>case_index>>num;
		for(int j=2; (j+1)*j <= 2*num;j++)
		{
			if((num-(j-1)*j/2) % j == 0)
			{
				count++;
			}
		}
		cout<<i<<" "<<count<<endl;
	}
	return 0;
}