Scientific Problem(数学，简单)_integer positive predecessors-优快云博客

本文介绍了一种算法，用于寻找一个整数n，该整数是给定整数x倍小于所有正整数前驱之和的解决方案。通过数学推导得出n大于2x，并给出了具体的实现代码。

Scientific Problem

Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Description

Once upon a time Professor Idioticideasinventor was travelling by train. Watching cheerless landscape outside the window, he decided to invent the theme of his new scientific work. All of a sudden a brilliant idea struck him: to develop an effective algorithm finding an integer number, which is x times less than the sum of all its integer positive predecessors, where number x is given. As far as he has no computer in the train, you have to solve this difficult problem.

Input

The first line of the input file contains an integer number x (1 ≤ x ≤ 10 ⁹).

Output

Output an integer number — the answer to the problem.

Sample Input

sample input	sample output
1	3

sample input	sample output
2	5

设答案为n，则nx < 1 +2 +3 + ... + n - 1 ,得到 n > 2x,所以n = 2x +1
AC CODE

//Memory: 935 KB 		Time: 31 MS
//Language: GNU CPP (MinGW, GCC 4) 		Result: Accepted

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define LL long long
#define MAXI 2147483647
#define MAXL 9223372036854775807
#define eps (1e-8)
#define dg(i) cout << "*" << i << endl;
using namespace std;

int main()
{
    LL x;
    while(scanf("%I64d", &x) != EOF)
    {
        printf("%I64d\n", 2 * x + 1);
    }
    return 0;
}