| The ? 1 ? 2 ? ... ? n = k problem |
Theproblem
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
TheInput
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
题意不累赘~
做法:
假设sum1 = a1 + a2 + a3 + ... + an + x >= k
而sum2 = a1 + a2 + a3 + ... + an - x = k
那么sum1 - sum2 = 2x
也就是说, 无论k的正负, 全把k当正数处理, 一直累加正数得到sum1 与 不按全当正数处理得到的sum2 相差的值是一个偶数(2x, 即负数的绝对值的两倍~)
故, 全部从1累加到n吧, 直到 (sum >= k && (sum - k) % 2 == 0)
AC代码:
#include<stdio.h>
int T;
int main() {
scanf("%d", &T);
while(T--) {
int k;
int sum = 0;
scanf("%d", &k);
if(k < 0)
k = (-1 * k);
for(int i = 1; ;i++) {
sum += i;
if(sum >= k && (sum-k) % 2 == 0) {
printf("%d\n", i);
break;
}
}
if(T)
printf("\n");
}
return 0;
}
