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\n",j); return 0; } } } return 0;
}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/49701671