UVA - 10025 ：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

212-3646397

Sample Output

72701

//首先不管k的正负性，因为一个负值存在那样的表达式的话一定相应的存在正值表达式。 其次，假设一直用的是加号定存在n使得sum=n*（n+1）/2>=k，而如果要变某个加号为减号也必定先满足上面的表达式，在这基础上由于改变一个加号为减号相当于在原表达式（+的表达式）的基础上减去了两倍的当前这个数，所以（sum-k）必须为偶数。那个第一个遇到的满足上面两个条件的n是不是就是要求的，答案是肯定的。因为(sum-k)肯定小于n*(n-1)/2。而这个差值肯定能由前面的n个数中的某几个做差得到。事实上在2*(2*(n-1)/2)内的数都可以由做差得到。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<cstdlib>#include<queue>#include<stack>#include<map>#include<vector>#include<algorithm>using namespace std;int main(){int T,t=0,k,i;scanf("%d",&T);while(T--){scanf("%d",&k);if(t>0)puts("");t++;k=abs(k);int sum=0;for(i=1;;i++){sum+=i;if(sum>=k&&(sum-k)%2==0){printf("%d\n",i);break;}}} return 0;}