[Ybtoj High-Efficiency Advanced 1.4] C. Worm Food Calculation [DFS]

analysis

Record the order of appearance of the letters (from right to left) and
enumerate the possible numbers for each letter

a is addend one, b is addend two, c is sum

When these three numbers have been filled in

1. If (a+b+w)%n!=c is not feasible,
   it is not feasible if there is a carry in the highest bit
2. When the right side is still unfilled
   if (a+b)%n!=c and (a+b+1)%n!=c, it is not feasible.
   If the highest bit has a carry, it is not feasible,
   otherwise the carry is assigned to -1

Upload code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

string s[10];
char f[30];
int n,t,p[30],ans[30],u[30];

bool check()
{
    
    
	int w=0;
	for(int i=n;i>0;i--)
	{
    
    
		int x=ans[s[1][i-1]-64];
		int y=ans[s[2][i-1]-64];
		int z=ans[s[3][i-1]-64];
		if(x!=-1&&y!=-1&&z!=-1)
		{
    
    
			if(w!=-1)
			{
    
    
				if((x+y+w)%n!=z) return false;
				if(i==1&&x+y+w>=n) return false;
				w=(x+y+w)/n;
			}
			else
			{
    
    
				if((x+y)%n!=z&&(x+y+1)%n!=z) return false;
				if(i==1&&x+y>n) return false;
			}
			
		}
		else w=-1;
	}
	return true;
}

int dfs(int x)
{
    
    
	if(x>n) return 1;
	for(int i=0;i<=n-1;i++)
	{
    
    
		if(!u[i])
		{
    
    
			ans[f[x]-64]=i;
			u[i]=1;
			if(check()&&dfs(x+1))
			{
    
    
				return 1;
			}
			u[i]=0;
			ans[f[x]-64]=-1;//回溯 
		}
	}
	return 0;
}

int main()
{
    
    
    memset(ans,-1,sizeof(ans));
	cin>>n;
	cin>>s[1];
	cin>>s[2];
	cin>>s[3];
	for(int i=n;i>=1;i--)
	{
    
    
		for(int j=1;j<=3;j++)
		{
    
    
			if(!p[s[j][i-1]-65])
			{
    
    
				p[s[j][i-1]-65]=1;
				f[++t]=s[j][i-1];
			}
		}
	}	
	dfs(1);
	for(int i=1;i<=n;i++)
	{
    
    
		cout<<ans[i]<<' ';
	}
	return 0;
}