uva10025 The ? 1 ? 2 ? ... ? n = k problem

10025 The ? 1 ? 2 ? … ? n = k 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

Input

The first line is the number of test cases, followed by a blank line. Each test case of the input contains an integer k (0 ≤|k|≤ 1000000000). Each test case will be separated by a single line.

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

首先预处理下，然后找到题目给出的计算结果位于的区间，由于数列中使用减号可以组合出任意的偶数，所以就查找与答案差的值为偶数的最小的那个数

#include<iostream>
#include<algorithm>
#include<map>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<cstring>
#include<string>
using namespace std;
const int maxn=54723;
typedef long long LL;

LL sum[maxn];

void work(LL n){
if(n==0){printf("3\n");return ;}
if(n<0)n=-n;
int high=lower_bound(sum,sum+maxn,n)-sum;
int ans=high;
while((sum[ans]-n)%2!=0)ans++;
printf("%d\n",ans);
}
int main(){
//    #ifndef ONLINE_JUDGE
//    freopen("Text//in.txt","r",stdin);
//    #endif // ONLINE_JUDGE
sum[0]=0;
for(int i=1;i<maxn;i++){
sum[i]=sum[i-1]+(LL)i;
}
int T;
scanf("%d",&T);
while(T--){
LL n;
scanf("%lld",&n);
work(n);
if(T>0)puts("");
}
return 0;
}

/*
Sample Input
2
12
-3646397
Sample Output
7
2701
*/