ac代码（~~很刺激的代码谁试谁知道~~）
```
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,b,c,d,e,f,g,h,i,j,ans,x=0;
cin>>ans;
for (a=1;a<=3;a++)
{
for (b=1;b<=3;b++)
{
for (c=1;c<=3;c++)
{
for (d=1;d<=3;d++)
{
for (e=1;e<=3;e++)
{
for (f=1;f<=3;f++)
{
for (g=1;g<=3;g++)
{
for(h=1;h<=3;h++)
{
for (i=1;i<=3;i++)
{
for (j=1;j<=3;j++)
{
if (a+b+c+d+e+f+g+h+i+j==ans)
{
x++;
}
}
}
}
}
}
}
}
}
}
}
cout<<x<<endl;
for (a=1;a<=3;a++)
{
for (b=1;b<=3;b++)
{
for (c=1;c<=3;c++)
{
for (d=1;d<=3;d++)
{
for (e=1;e<=3;e++)
{
for (f=1;f<=3;f++)
{
for (g=1;g<=3;g++)
{
for(h=1;h<=3;h++)
{
for (i=1;i<=3;i++)
{
for (j=1;j<=3;j++)
{
if (a+b+c+d+e+f+g+h+i+j==ans)
{
cout<<a<<" ";
cout<<b<<" ";
cout<<c<<" ";
cout<<d<<" ";
cout<<e<<" ";
cout<<f<<" ";
cout<<g<<" ";
cout<<h<<" ";
cout<<i<<" ";
cout<<j<<endl;
}
}
}
}
}
}
}
}
}
}
}
return 0;
}
```
by Alpha丶 @ 2018-11-11 18:05:19
~~~
#include<iostream>
#include<cmath>
using namespace std;
int now[10],zs;
void print()
{
for(int i=0;i<10;i++)cout<<now[i]<<" ";
cout<<endl;
}
void ans_z(int n,int i)
{
if(i==9){now[9]=n;zs++;return;}
for(int j=max(n-27+3*i,1);j<=min(3,n-9+i);j++)
{
now[i]=j;
ans_z(n-j,i+1);
}
return;
}
void ans_p(int n,int i)
{
if(i==9){now[9]=n;print();return;}
for(int j=max(n-27+3*i,1);j<=min(3,n-9+i);j++)
{
now[i]=j;
ans_p(n-j,i+1);
}
return;
}
int main()
{
int n;
cin>>n;
ans_z(n,0);
cout<<zs<<endl;
ans_p(n,0);
return 0;
}
~~~
by ttklwxx @ 2018-11-11 18:05:39
至于dfs我也很无能为力啊
（~~毕竟只是新手村的题目~~）
by Alpha丶 @ 2018-11-11 18:06:11
退役后第一次手写一下核心代码，CE别怪我
~~方案什么的就自己预处理去吧~~
```cpp
int use[1000], n;
inline void dfs(int now) {
if(now == 11) {
int all = 0;
for(int i = 1; i <= 10; ++i) all += use[i];
if(all == n) {
for(int i = 1; i <= 10; ++i) printf("%d ", use[i]);
cout << endl;
}
return ;
}
for(int i = 1; i <= 3; ++i) {
use[now] = i;
dfs(now+1);
}
return ;
}
```
by Fraction @ 2018-11-11 18:13:44
```cpp
#include<iostream>
using namespace std;
int a[11],x=0,n,s=0,ans[11][10001];
void so(int m)
{
for(int i=1;i<=3;i++)
if(m<=10)
{
x+=i;
a[m]=i;
if(m==10&&x==n)
{
s++;
for(int j=1;j<=10;j++)
ans[j][s]=a[j];
}
else so(m+1);
x-=i;
}
}
int main()
{
cin>>n;
so(1);
cout<<s<<endl;
for(int i=1;i<=s;i++)
{
for(int j=1;j<=10;j++) cout<<ans[j][i]<<" ";
cout<<endl;
}
return 0;
}
```
by csdfret @ 2018-11-11 18:15:25
入门难度的dfs
by FCBM71 @ 2018-11-11 18:23:29