[sicily]1209. Sequence Sum Possibi

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

简单的数论题。给定一个数，找出有多少组连续数字之和为该数。利用等差数列公式，枚举项数。设有 i 项，为 l,l+1,l+2....l+i-1; 则，(2l + i-1 ) * i = 2n ; 即：2l*i + (i-1)*i = 2n，由于 l 大于0的整数，则，可判断出 i 的范围以及 i 能否成为项数的条件。具体代码如下：

#include <iostream>using namespace std;int main(){    int t;    cin>>t;    while(t--)    {        int count = 0;        int index, num;        cin>>index>>num;        for(int i=2; i*i-i < 2*num; i++)        {            if( (2*num - i*(i-1)) % (2*i) == 0)                count++;        }        cout<<index<<" "<<count<<endl;            }     //system("pause");    return 0;}