Codeforces 964A Splits

最近写了几场cf的题， 各种浑水摸鱼过一道就跑导致分数低到令人发指，也并不知道自己都写了些什么， 遂总结一下这几次的题目回忆一下。

                                                                                                                                                                      以上。

Description：

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 : $[4,4]$ $[3,3,2]$$[2,2,1,1,1,1]$. The following sequences aren't splits of : $[1,7]$, $[11,-3$$]$. 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 , the weight of the split is 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 ($1\le n\le {10}^9$). Output Output one integer — the answer to the problem. Examples Input 7 Output 4 Input 8 Output 5 Input 9 Output 5 Note In the first sample, there are following possible weights of splits of : Weight 1: [] Weight 2: [, , 1] Weight 3: [, , $\text{2}$, 1] Weight 7: [, , , , , , ]

题目大意：

给一个整数n，统计n有多少种不同题目所要求的拆分。 所谓拆分指某组数和为n， 且它们按照降序排序后第一个元素的个数。统计有多少组满足条件的数组（第一个元素个数必须不同）。

解题思路：

逆向思维， 我们从底层开始往上推， 从全1开始，每次合并两个1，直到全部为2。 那么些写现在我们其实可以知道这道题只用计算2的个数即可。

代码：

#include <iostream>
#include <sstream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <utility>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>

using namespace std;

/*tools:
*ios::sync_with_stdio(false);
*freopen("input.txt", "r", stdin);
*/

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int dir[9][2] = { 0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1, 0, 0 };
const int inf = 0x3f3f3f;
const double pi = acos(-1.0);
const int mod = (ll) 1e9 + 7;
const int Max = (int) 1e5 + 51;
const ld eps = 1e-10;

int n;
int num;

int main() {
    while (~scanf("%d", &n)) {
        num = n / 2;
        printf("%d\n", num + 1);
    }

    return 0;
}