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

題目大意：輸入一個整數n，問總共有多少個連續序列之和爲這個數。
解題方法：若是直接從0開始遍歷依次確定超時，在這裏這個序列確定爲一個公差爲1的等差數列，假設首項爲a1,長度爲i,若是知足條件，則n = a1 * i + i * (i - 1) / 2;
即n -i * (i - 1) / 2 = a1 * i;也就是說n的值爲長度爲i，首項爲a1的等差數列之和，因此只要判斷（n -i * (i - 1) / 2） % i是否爲0便可，固然長度i有一個範圍，假設a1爲最小值1,那麼長度i確定爲最大值，n = i + i * (i - 1) / 2，即n = i * (i + 1) / 2，因此i的最大值不會超過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;
}