UVA 10025 - The ? 1 ? 2 ? ... ? n = k problem（数学）

The ? 1 ? 2 ? ... ? n = k problem

The problem

Given the following formula, one can set operators '+' or '-' instead of each '?', in order to obtain a given k
? 1 ? 2 ? ... ? n = k

For example: to obtain k = 12 , the expression to be used will be:
- 1 + 2 + 3 + 4 + 5 + 6 - 7 = 12 
with n = 7

The Input

The first line is the number of test cases, followed by a blank line.

Each test case of the input contains integer k (0<=|k|<=1000000000).

Each test case will be separated by a single line.

The Output

For each test case, your program should print the minimal possible n (1<=n) to obtain k with the above formula.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

2

12

-3646397

Sample Output

7

2701

---

Alex Gevak
September 15, 2000 (Revised 4-10-00, Antonio Sanchez)

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给你一个整数k，问你至少需要多少个从1开始的连续整数（加减都可以）得到k（k的正负不影响结果，可以把所有的正负调换结果相同）

先预处理一下，把从1开始到i的连续整数和保存在sum[i]里，现在所有整数前面的符号都为正。对于输入的k，如果sum[i]-k为偶数的话，则一定可以在连续整数中找到一个数把前面的符号改为负，使sum结果等于k（若sum[i]-k=0，则符号均为正即可），输出i；如果sum[i]-k为奇数则i++直到sum[i]-k为偶数输出i即可

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int sum[55555]={0},p,n;

int main()
{
    for(int i=1;sum[i-1]<1000000000;i++)
    {
        sum[i]=sum[i-1]+i;
    }
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        if(n==0)
        {
            printf("3\n");
            if(t) printf("\n");
            continue;
        }
        if(n<0) n=fabs(n);
        for(int i=0;;i++)
        {
            if(sum[i]>=n)
            {
                p=i;
                break;
            }
        }
        if((n-sum[p])%2==0) printf("%d\n",p);
        else
        {
            p++;
            while((n-sum[p])%2)
            { 
                p++;
            }
            printf("%d\n",p);
        }
        if(t) printf("\n");
    }
    return 0;
}