Sum

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描述

Consider the natural numbers from 1 to N. By associating to each number a sign (+ or -) and calculating the value of this expression we obtain a sum S. The problem is to determine for a given sum S the minimum number N for which we can obtain S by associating signs for all numbers between 1 to N. 

For a given S, find out the minimum value N in order to obtain S according to the conditions of the problem. 

输入

The input consists N test cases.
The only line of every test cases contains a positive integer S (0< S <= 100000) which represents the sum to be obtained.
A zero terminate the input.
The number of test cases is less than 100000.

输出

The output will contain the minimum number N for which the sum S can be obtained.

样例输入

3
12
0

样例输出

2
7

思路：

给定一个数n，在1+2+3+4+…+k中，求随意改变加减号能使其和（差）正好为n的k为多少，使得k最小。解题思路：每次累加i：如果sum小于n则无论如何也不可能达到题意；如果sum正好等于n则累加到这个k正好为答案；如果sum大于n时，则需要把前面的加号改为减号：如果改为-1，结果就减2，如果改为-2，结果就减4，以此类推，可以看出只要sum-n为偶数，则此时可以改sum结果为n，得答案。
 
代码：

#include<stdio.h>
int main()
{
	int n,i,t;
	while(scanf("%d",&n)&&n)
	{
		i=1;
		t=1;
		while(t!=n)
		{
			i++;
			t=i*(i+1)/2;
			if(t>n)
			{
				if((t-n)%2==0)
					t=n;
			}
		}
		printf("%d\n",i);
	}
	return 0;
}

Sum

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原文地址：http://blog.csdn.net/qq_16997551/article/details/47210529