Sicily1209. Sequence Sum Possibi题解

1. 题目描述

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.

【翻译过来】：给了一个数字，将其分解为至少2个连续整数的和的形式，那么一共有多少种方法？

2. 样例

【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.

Input：
7
1 6
2 9
3 8
4 1800
5 987654321
6 987654323
7 987654325
Output：
1 1
2 2
3 0
4 8
5 17
6 1
7 23

3. 分析

拿到题目，乍一分析，已为用分治法的思想解决，即：把当前一个大的数字，将其除以2，不断分析规模较小的数字情况，从而加起来。应该用到递归的策略，但是没有找到合适的算法，于是继续分析，发现了数学规律。

【公式1】：

假设需要计算的目标数量为value，一共有以正整数n开始的j个连续的数相加，那么可以得到假设：

value=n+(n+1)+(n+2)+……+(n+j−1)

value=(n+n+j−1)∗j/2

2∗value=(n+n+j−1)∗j

n≥1=>(2n−1)≥1

2∗value≥(j+1)∗j

2∗value≥j2

因此我们可以知道，满足条件的k，即连续k个正整数相加等于N，那么k和N满足这样的关系。

【公式2】：

接下来，继续推导，由上一个推导：

value=n+(n+1)+(n+2)+……+(n+j−1)

value=j∗n+1+2+3+……+j−1

value=j∗n+(1+j−1)∗(j−1)/2

value−(1+j−1)∗(j−1)/2=j∗n

value−j∗(j−1)/2=j∗n

j与n均为正整数，则value−j∗(j−1)/2是j的倍数。

所以，一个循环遍历，直到满足公式1得到的条件：2∗value≥j2，代表这个数value理论上最多可以由这些整数相加而得。

在这个循环下，判断满足题意的条件，即公式2：value−j∗(j−1)/2是j的倍数，每满足一个这个条件的j值，计数器就记录一次这样的分法（代表此时由j个正整数相加得到），返回计数器值即可。

4. 源码

/*
* main.cpp
*
*  Created on: 2017年6月8日
*  Author: liboyang
*/
#include <iostream>
using namespace std;

int calculate(int value);
int main() {
int N;
int number, value;
cin >> N;
int counter = N;
while(counter > 0) {
counter--;
cin >> number >> value;
int result = calculate(value);
cout << number << " " << result << endl;
}
return 0;
}
int calculate(int value) {
int count = 0;
for (int j = 2; j * j <= 2*value; j++) {
if((value - (j-1)*j/2) % j == 0) {
count++;
}
}
return count;
}

5. 心得

这道题是纯数学题，总感觉算法期末考试有点方= =