Sicily 1209 Sequence Sum Possibi

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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

1 1
2 2
3 0
4 8
5 17
6 1
7 23

Solution

题目大意是给定n，求n可以由等于多少段连续自然数之和。

数学题，我一开始想的是直接穷举起点和终点，但是无论怎么优化都是会超时的

然后就学到枚举项数，反过来算起点是不是整数来判断是不是一个合法的解，注意项数的上限。

#include <iostream>

using namespace std;

int main()
{
int t;

cin >> t;
while ( t -- )
{
int i, j, t_case, n, count = 0;

cin >> t_case >> n;
for ( i = 2; i * i - i < 2 * n; i ++ )//项数枚举
{
if ( (n - i * ( i - 1 ) / 2) % i == 0 ) count ++;
}
cout << t_case << ' ' << count << '\n';
}

return 0;
}
//两个限制条件，整除和枚举的不是起点和终点，而是项数，通过项数反算起点是否为整数