递归思路

#include<iostream>
using namespace std;

int ans=1,n;
void solve(int m){
    if(m==1){
        //ans++;如果在这里记录答案，只是结果中包含1的答案，不完整
        return ;
    }
    int x=m/2;
    for(int i=x; i>=1; i--)//对应题目加上不超过原数一半的全部自然数
    {
            ans++;
          solve(i); 
    }
}
int main(){
    cin>>n;
    solve(n);
    cout<<ans;
    return 0;
} 

#include<iostream>
using namespace std;
int r=0;//定义一个全局变量record的缩写，记录个数
inline void digui(int n)//这个地方inline应该不起作用。感觉有 就加了
{
if(n/2==0)//判断是不是一半
{
return;
}
for(int i=1;i<=n/2;i++)//枚举每一种
{//用树的图样子更好理解.
r++;
digui(i);
}
return ;
}
int main(){
int n;
cin>>n;
r++;
digui(n);
cout<<r<<endl;
return 0;
}//对于这个题，本人希望你能够用笔写一下 就能找出关系虽然我手机上过不了，但你网上就能。

#include<bits/stdc++.h>
using namespace std;
int h[1001];
int fun(int n) {
int i;
h[1]=1;
for(i=2; i<=n; i++) {
if(i%2==1)h[i]=h[i-1];
else h[i]=h[i-1]+h[i/2];
}
return h[n];
}
int main() {
int n;
while(scanf("%d",&n)!=EOF) {
printf("%d\n",fun(n));
}
return 0;
}
已经AC

#include<bits/stdc++.h>
using namespace std;
int n;
int f[1001];//存每一位数的种类
int main(){
cin>>n;
for(int i=1;i<=n;i++){ //1-n的递推
for(int j=1;j<=i/2;j++){
f[i]+=f[j]; //每一位叠加
}
f[i]++; //加上本身
}
cout<<f[n];//输出n的种类
return 0;
}

#include<bits/stdc++.h>
int num=0;
int find(int n){
    ++num;
    for(int i=n/2;i>=1;--i) find(i);
}
int main(){
    int n;
    scanf("%d",&n);
    find(n);
    printf("%d",num);
}

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <bitset>
#include <cassert>
#include <map>
#include <string>
#include <sstream>
#include <ctime>
using namespace std;
int x;
int dfs(int n)
{
int i,a=0;
if(n==1)
{
return 0;
}
else
{
a+=n/2;
for(i=1;i<=n/2;i++)
{
a+=dfs(i);
}

}
return a;
}
int main()
{
scanf("%d",&x);
printf("%d",dfs(x)+1);
return 0;
}

递推即可

#include<iostream>  
using namespace std;  
int main()  
{  
    int a[1001],n;  
    cin>>n;  
    a[1]=1;  
    for(int s=2;s<=n;s++)  
    {  
        a[s]=0;  
        for(int i=1;i<=s/2;i++)  
        {  
            a[s]+=a[i];  
        }  
        a[s]++;  
    }  
    cout<<a[n];  
    return 0;  
}  

其实题意不太清，简述：将n作出操作后会形成新的数列，实际上数的是能构成多少合法数列，且过程中可任意停止操作，即任意不同的前缀数列都可称为新数列。则用dp来做，f[i]表示以 i 这个数字为开头可生成的所有新数列.
则可递推f[i]=f[1]+f[2]+...+f[i/2];有兴趣的同学可以用前缀和优化一小下下。。
下面是0msAC代码：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int f[1005],n;
int main()
{
scanf("%d",&n);
f[1]=1;
for(int i=2;i<=n;i++){
f[i]=1;
for(int j=1;j<=(i/2);j++){
f[i]+=f[j];
}
}
cout<<f[n];
}

#include<iostream>
using namespace std;
const int maxn = 1000 + 5;
int n, a[maxn]={0};
int dfs(int t){
    int h= t / 2;
    if(a[h]) return a[h];
    a[h] = 1;
    for(int i = 1; i <= h; i++) a[h] += dfs(i);
    return a[h];
}
int main(){
    cin >> n;
    cout << dfs(n);
    return 0;

}

#include <cstdio>

int dfs(int x){
int half,sum=1;
if(x%2==0)
half=x/2;
else
half=(x-1)/2;
for(int i=1;i<=half;i++)
sum+=dfs(i);
return sum;
}

int main(){
int n;
scanf("%d",&n);
int total=dfs(n);
printf("%d",total);
return 0;
}

var n,i:integer;
f:array[0..1001] of longint;
begin
readln(n);
f[0]:=1;
for i:=1 to n do
if i mod 2=1 then f[i]:=f[i-1]
else
f[i]:=f[i-1]+f[i div 2];
writeln(f[n]);
end.
So easy!

此题太水，随便就过了
var
s,n:longint;
procedure dt(m:longint);
begin
if m>0 then while m>0 do
begin
inc(s);
dt(m div 2);
dec(m);
end;
end;
begin
readln(n);
s:=1;
dt(n div 2);
write(s);
end.

C：
#include <stdio.h>
int main() {
long long A[1001];
int n;
int i;
A[1] = 1;
scanf("%d", &n);
for(i=2; i<=n; i++) {
if(i%2 == 0) A[i] = A[i/2] + A[i-1];
else A[i] = A[i-1];
}
printf("%I64d", A[n]);
return 0;
}

var
a:array[1..1000]of longint;
i,n:longint;
begin
a[1]:=1;
readln(n);
for i:=2 to n do
if odd(i) then a[i]:=a[i-1]
else a[i]:=a[i-1]+a[i div 2];
writeln(a[n]);
end.

晕_(:з」∠)_这都能A
```java
import java.util.*;
import java.math.*;

public class Main {
static BigInteger count = BigInteger.ZERO;

static void checkAndRun(int x) {
if (x == 0) {
count = BigInteger.ONE;
} else {
loop(x);
}
}

static void loop(int x) {
if (x > 0) {
count = count.add(BigInteger.ONE);
} else {
return;
}
for (int i = x / 2; i >= 0; i--) {
loop(i);
}
}

public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
checkAndRun(scanner.nextInt());
System.out.println(count.toString());
scanner.close();
}
}
```

记忆化搜索
```pascal
var n,i:longint;
a:array[0..1001] of longint;
function cal(k:longint):longint;
var i,ans:longint;
begin
if k=0 then exit;
if a[k]<>-1 then exit(a[k]);
ans:=k div 2;
for i:=1 to k div 2 do begin
inc(ans,cal(i));
end;
a[k]:=ans;
cal:=ans;
end;

begin
fillchar(a,sizeof(a),-1);
read(n);
write(cal(n)+1);
end.
```

暴力出奇迹

递推。规律。
```
/* ***********************************************
Author :guanjun
Created Time :2016/2/19 14:25:19
File Name :vijosp1130.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
int x,y;
};
struct cmp{
bool operator()(Node a,Node b){
if(a.x==b.x) return a.y> b.y;
return a.x>b.x;
}
};

bool cmp(int a,int b){
return a>b;
}
int main()
{
#ifndef ONLINE_JUDGE
//freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
ll f[1100],n;
f[1]=1;
f[2]=f[1]+1;
f[3]=f[1]+1;
f[4]=f[1]+f[2]+1;
f[5]=f[1]+f[2]+1;
f[6]=f[1]+f[2]+f[3]+1;
for(int i=4;i<=1001;i++){
int x=i/2;
f[i]=0;
for(int j=1;j<=x;j++){
f[i]+=f[j];
}
f[i]++;
}
while(cin>>n){
cout<<f[n]<<endl;
}
return 0;
}

var
n,ans:longint;
procedure go(t:longint);
var
i:longint;
begin
if t>0 then
for i:=1 to t do
inc(ans);
for i:=1 to t do
go(i div 2);
end;
begin
readln(n);
go(n div 2);
writeln(ans+1);
end.

program numbercount;

var n,ans:longint;

procedure f(x:longint);
var n:longint;
begin
n:=x div 2;
while n>0 do begin
inc(ans);
f(n);
dec(n);
end;
end;

begin
readln(n);
ans:=1;
f(n);
writeln(ans);
end.