Codeforces Round #475 (Div. 2) A. Splits

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本文介绍了一种算法问题，即将一个整数拆分成非递增的序列，并计算这些序列中特定元素的出现次数（权值）。文章提供了一个简单的C++实现方案，通过将数字尽可能多地拆分为2来快速找出所有可能的权值。

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Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$ .

For example, the following sequences are splits of $8$ : $[4, 4]$ , $[3, 3, 2]$ , $[2, 2, 1, 1, 1, 1]$ , $[5, 2, 1]$ .

The following sequences aren't splits of $8$ : $[1, 7]$ , $[5, 4]$ , $[11, - 3]$ , $[1, 1, 4, 1, 1]$ .

The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$ , the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$ .

For a given $n$ , find out the number of different weights of its splits.

Input

The first line contains one integer $n$ ( $1 \leq n \leq 10^{9}$ ).

Output

Output one integer — the answer to the problem.

Examples
inputCopy
7
outputCopy
4
inputCopy
8
outputCopy
5
inputCopy
9
outputCopy
5

Note

In the first sample, there are following possible weights of splits of $7$ :

Weight 1: [ $7$ ]

Weight 2: [ $3$ , $3$ , 1]

Weight 3: [ $2$ , $2$ , $2$ , 1]

Weight 7: [ $1$ , $1$ , $1$ , $1$ , $1$ , $1$ , $1$ ]

题意：把一个数拆成非增的序列，拆分成的序列的权值是序列第一个数字的个数。

思路：任意一个数拆分成尽量多的2，比如7可以拆分成[2,2,2,1]，这是非全1的时候的最大权值，然后依次把2拆分成1就能使权值变小，即[2,2,1,1,1] [2,1,1,1,1,1]，再加上一个全1就是所求。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
int main()
{
    int n,i,ans;
    while(scanf("%d",&n)!=EOF)
    {

        printf("%d\n",(n/2)+1);
    }
    return 0;
}