uva 10025 The ? 1 ? 2 ? ... ? n = k problem（算式规律）-优快云博客

本文链接：https://blog.youkuaiyun.com/keshuai19940722/article/details/9664075

该博客介绍了如何解决UVA 10025问题，即通过设置加减运算符使得等式? 1 ? 2 ? ... ? n = k成立。解题思路涉及寻找最小的n值，确保表达式的和大于n，并且考虑通过减去偶数来调整结果。关键在于保证和与n的奇偶性一致。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

7

2701

题目大意：给出一个数，判断它需要从1~m（可用+、-）填充的算式求得。求m

解题思路：首先， 1~m的和一定要 > n(如果全+都无法表示的话，更别说有-）其次，找到一个m,使得1~m的和>n之后，因为sum >= n, 所以要减掉一个数，比如在数字k前面加个-号，相当于sum - 2 * k，也就是说每次减掉只能是偶数，那么就要求sum % 2 == n % 2.(负数同理）

#include<stdio.h>

int main(){
	int t, n, sum, i;
	scanf("%d", &t);
	while (t--){
		scanf("%d", &n);
		if (n < 0)
			n = -n;
		for (sum = 0, i = 1; ; i++){
			sum += i;
			if (sum >= n && (sum - n) % 2 == 0){
				printf("%d\n", i);
				break;
			}
		}
		if (t)
			printf("\n");
	}
	return 0;}