[Ybtoj High-Efficiency Advanced 1.4] [Deep Search] Worm Eater
topic
Problem-solving ideas
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
- If (a+b+w)%n!=c is not feasible,
it is not feasible if there is a carry in the highest bit - When the number on the right is not filled,
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
Code
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
string s[5];
char q[30];
int n,t,p[30],ans[30],use[30];
bool check()
{
int w=0;
for (int i=n;i>0;i--)
{
int x=ans[s[1][i-1]-64],y=ans[s[2][i-1]-64],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;
}
bool dfs(int x)
{
if (x>n) return true;
for (int i=0;i<n;i++)
if (!use[i])
{
ans[q[x]-64]=i;
use[i]=1;
if (check()&&dfs(x+1))
return true;
use[i]=0;
ans[q[x]-64]=-1;
}
return false;
}
int main()
{
memset(ans,-1,sizeof(ans));
scanf("%d",&n);
cin>>s[1];
cin>>s[2];
cin>>s[3];
for (int i=n;i>0;i--)
for (int j=1;j<=3;j++)
if (!p[s[j][i-1]-65])
{
p[s[j][i-1]-65]=1;
q[++t]=s[j][i-1];
}
dfs(1);
for (int i=1;i<=n;i++)
printf("%d ",ans[i]);
return 0;
}