[HNU 10052] Sum

原创于 2020-12-19 22:11:05 发布 · 344 阅读

1 ·

CC 4.0 BY-SA版权

算法同时被 2 个专栏收录

42 篇文章

订阅专栏

每日一题

25 篇文章

订阅专栏

博客介绍了一道AC题的解法。该题若需输出可行序列是背包问题，但本题无需输出序列。对于构造的sum序列，将一个“+”变为“-”会减少一个偶数，只需从n开始向后判断与题目给定的n做差是否为偶数。

Sum

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 1451, Accepted users: 1296

Problem 10052 : No special judgement

Problem description

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.

Input

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

Output

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

Sample Input

Sample Output

Judge Tips

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

Problem Source

ROI 2002

AC:

#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <cmath>
using namespace std;

int main()
{
    int n;
    scanf("%d", &n);
    int i = floor(-1 + sqrt(1 + 8 * n)) / 2, sum = i * (1 + i) / 2;
    for (;;)
    {
        if (sum >= n && !((sum - n) & 1))
        {
            printf("%d", i);
            break;
        }
        sum += (++i);
    }
    return 0;
}

这个题还是比较巧妙的，如果需要输出可行的序列，就是一个背包问题了，但是这个题，并不需要输出序列，对于构造的sum序列，每次将一个+变为-，就会减少一个偶数（在原来的值上减去两倍的变符号的值）。所以整体上只需要从n开始向后判断是否和题目给定的n做差得到偶数即可。