NUC1196 Sum【水题+数学题】-优快云博客

本文探讨了一个数学问题：如何找到最小的自然数N，使得从1到N的序列通过分配正负号能得到特定的目标和S。文章提供了一种有效的算法，并附带了C++实现代码。

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Sum

时间限制: 1000ms 内存限制: 65536KB

通过次数: 1总提交次数: 1

问题描述

Consider the natural numbers from 1 to N. By associating to each number a sign (+ or -) and calculating the value of this expression we obtain a sum S. The problem is to determine for a given sum S the minimum number N for which we can obtain S by associating signs for all numbers between 1 to N.

For a given S, find out the minimum value N in order to obtain S according to the conditions of the problem.

输入描述

The only line contains in the first line a positive integer S (0 < S ≤ 100000) which represents the sum to be obtained.

输出描述

The output will contain the minimum number N for which the sum S can be obtained.

样例输入

样例输出

来源

PKU

提示

The sum 12 can be obtained from at least 7 terms in the following way: 12 = -1+2+3+4+5+6-7.

问题分析：（略）

这个问题和《POJ1844 Sum【水题+数学题】》是同一个问题，代码直接用就AC了。

程序说明：参见参考链接。

参考链接：POJ1844 Sum【水题+数学题】

题记：程序做多了，不定哪天遇见似曾相识的。

AC的C++程序如下：

/* POJ1844 Sum */

#include <iostream>
#include <math.h>

using namespace std;

int main()
{
    int s;

    while(cin >> s) {
        for(int i=sqrt(s); ;i++) {
            int sn = (i + 1) * i / 2;
            if(sn >= s && (sn - s) % 2 == 0) {
                cout << i << endl;
                break;
            }
        }
    }

    return 0;
}

转载于:https://www.cnblogs.com/tigerisland/p/7563666.html