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
*/