Codeforces 964A Splits

最近寫了幾場cf的題， 各種渾水摸魚過一道就跑導致分數低到令人髮指，也並不知道自己都寫了些什麼， 遂總結一下這幾次的題目回憶一下。

                                                                                                                                                                      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: [, , , , , , ]

題目大意：

給一個整數n，統計n有多少種不同題目所要求的拆分。 所謂拆分指某組數和爲n， 且它們按照降序排序後第一個元素的個數。統計有多少組滿足條件的數組（第一個元素個數必須不同）。

解題思路：

逆向思維， 我們從底層開始往上推， 從全1開始，每次合併兩個1，直到全部爲2。 那麼些寫現在我們其實可以知道這道題只用計算2的個數即可。

代碼：

#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;
}