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

   比较简单的题目， 首先先找出满足1+...+n的和Sn大于或等于k的最小的n，用二分法找出，特殊情况k=0直接输出。第二个就是判断Sn的奇偶性是否与k相等，因为1~n中某一个数字m符号改为-号是，相当于n项和减小偶数2*m，所以Sn必定与k奇偶性相同。

#include <iostream>using namespace std;int main(){    int t, k;    cin >> t;    while ( t--){        cin >> k;        if ( k<0) k = -1*k;        if ( !k){            cout << 3 << endl;            if( t) cout << endl;            continue;        }        int a = 0, b = 50000,n;        int Sn;        while ( a <= b){                       //二分查找n的值            n = ( a + b )/2;            Sn = (n*n+n)/2;            if(  Sn-n < k &&  Sn >= k) break; //Sn>=k            else if ( k <= Sn - n ) b = n;            else if ( k > Sn ) a = n + 1;        }        if( (Sn&1) != (k&1)){   //判断奇偶性是否相同，若不同Sn还需要加一个奇数            if( n&1) n+=2;      // n为奇数，则n+1为偶数，需要n+2            else n++;           // n为偶数，则n+1为奇数        }        cout << n << endl;        if( t) cout << endl;    }}