POJ 2853 Sequence Sum Possibilities

Sequence Sum Possibilities

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5537		Accepted: 3641

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 2³¹ (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

题目大意：输入一个整数n，问总共有多少个连续序列之和为这个数。
解题方法：若是直接从0开始遍历依次确定超时，在这里这个序列确定为一个公差为1的等差数列，假设首项为a1,长度为i,若是知足条件，则n = a1 * i + i * (i - 1) / 2;
即n -i * (i - 1) / 2 = a1 * i;也就是说n的值为长度为i，首项为a1的等差数列之和，因此只要判断（n -i * (i - 1) / 2） % i是否为0便可，固然长度i有一个范围，假设a1为最小值1,那么长度i确定为最大值，n = i + i * (i - 1) / 2，即n = i * (i + 1) / 2，因此i的最大值不会超过sqrt(n * 2.0)。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;

int main()
{
    int nCase, index, n;
    scanf("%d", &nCase);
    while (nCase--)
    {
        int ans = 0;
        scanf("%d%d", &index, &n);
        for (int i = 2; i <= sqrt((double)n * 2.0); i++)
        {
            if ((n - i * (i - 1) / 2) % i == 0)
            {
                ans++;
            }
        }
        printf("%d %d\n", index, ans);
    }
    return 0;
}