Sequence Sum Possibi

描述

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.

输入

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 231 (so will fit in a 32-bit integer).

输出

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.

样例输入

7

1 6

2 9

3 8

4 1800

5 987654321

6 987654323

7 987654325

样例输出

1 1

2 2

3 0

4 8

5 17

6 1

7 23

纯数学题。

1.

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

2.

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
 

代码：

#include<iostream>
using namespace std;
int main()
{
int n;
cin>>n;
while(n--)
{
int a,b,m=0;
cin>>a>>b;
for(int i=2;i*(i+1)<=b*2;i++)
{
if((b-(i*(i-1)/2))%i==0)
m++;
}
cout<<a<<" "<<m<<endl;
}
return 0;
}