POJ 1844 Sum

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

12
Sample Output

7

第一次知道了，打表法原来也是要消耗时间的，只是相对少些；
还有，用scanf输入比用cin输入要节约时间；scanf是格式化输入，printf是格式化输出。
cin是输入流，cout是输出流。效率稍低，但书写简便。
格式化输出效率比较高，但是写代码麻烦。
流输出操作效率稍低，但书写简便。
cout之所以效率低，是先把要输出的东西存入缓冲区，再输出，导致效率降低。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXX 100010
using namespace std;
int a[MAXX];
void aa()
{
    a[0]=0;
    for(int j=1; j<=MAXX; j++)
    {
        a[j]=a[j-1]+j;
    }
}
int main()
{
    aa();
    int n;
    scanf("%d",&n);
    int k;
    for(int j=1;j<n; j++)
    {
        if(a[j]>=n)
        {
            k=a[j]-n;
            if(k%2==0)
            {
                printf("%d
",j);
                return 0;
            }
        }
    }
    return 0;
}