# Poj Solution 2842

```http://poj.org/problem?id=2842

#include <stdio.h>

int an[10], bn[10];
int sa[10], sb[10];
int p[10], n, m, g, h;
int S[500000], T[500000];

int get_T( int *a ) {
int index = 0, i;
for( i=0; i<n; i++ )
index += a[i]*sa[i];
return T[ index ];
}

int get_S( int *a ) {
int index = 0, i;
for( i=0; i<n; i++ )
index += (a[i]+p[i])*sb[i];
return S[ index ];
}

bool check( ) {
int a[10] = { 0 }, i, k;
for( i=0; i<m; i++ ) {
if( get_T( a ) != get_S( a ) )
return false;
for( k=n-1; a[k] == an[k]-1 && k; k-- )
;
a[k]++;
while( k++ < n-1 )
a[k] = 0;
}
return true;
}

void search( ) {
int i, k;
for( i=0; i<g; i++ ) {
if( check( ) )
return;
for( k=n-1; p[k] == bn[k]-an[k] && k; k-- )
;
p[k]++;
while( k++ < n-1 )
p[k] = 0;
}
return;
}

int main( ) {
int i;
g = m = h = 1;

scanf( "%d", &n );

for( i=0; i<n; i++ )
scanf( "%d", bn+i );
for( i=0; i<n; i++ ) {
scanf( "%d", an+i );
m *= an[i];
g *= bn[i]-an[i]+1;
h *= bn[i];
}

for( i=0; i<h; i++ )
scanf( "%d", S+i );
for( i=0; i<m; i++ )
scanf( "%d", T+i );

sb[n-1] = sa[n-1] = 1;

for( i=n-2; i>=0; i-- ) {
sa[i]=an[i+1]*sa[i+1];
sb[i]=bn[i+1]*sb[i+1];
}

search( );
for( i=0; i<n; i++ )
printf( "%d ", p[i]+1 );
printf( "n" );
return 0;
}
```
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