UVa 10025 The ? 1 ? 2 ? ... ? n = k problem

 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

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

 

题目的意思是说，在从1~n这n个数之间，通过添加+、-号使表达式的值为k，求这样的n最小为多少

通过列举前几个n的值可以组合出的k，很容易发现规律：

n=1 :　k=1,-1

n=2 :   k=3,-3,1,-1

n=3 :   k=6,-6,4,-4,2,-2,0

n=4 :   k=10,-10,8,-8,6,-6,4,-4,2,-2,0

n=5 :   k=15,-15,13,-13,11,-11,9,-9,7,-7,5,-5,3,-3,1,-1,0

……     ……

可以看出，当(n+1)/2向下取整为奇数时，可以组合出不大于n(n+1)/2的所有奇数

当(n+1)/2向下取整为偶数时，可以组合出不大于n(n+1)/2的所有偶数

按照以上规律写程序求解n即可

 

WA了3次，第一次发现了把浮点数转化为它向上取整的整数时写的有问题，第二次发现当k=0时需要特殊判断，第三次发现在解二次方程求出的那个表达式中有8*k这个会溢出int的范围，要强制类型转换一下

 

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

using namespace std;

int main()
{
    int kase;

    scanf("%d",&kase);

    while(kase--)
    {
        int k;

        scanf("%d",&k);

        if(k<0)
            k=-k;

        double temp=(sqrt(8*((long long)k)+1)-1)/2;
        int n=temp;

        if(fabs(temp-n)>1e-9)
            n++;

        bool even=(k%2==0);

        if(even)
            while(((n+1)/2)%2)
                n++;
        else
            while(((n+1)/2)%2==0)
                n++;

        printf("%d\n",k==0?3:n);
        if(kase)
            printf("\n");
    }

    return 0;
}