【链接】 我是链接,点我呀:)
【题意】
【题解】
```cpp
/*
for a[1] from 1~9
1*1=1 2*1=2 3*1=3
1*2=2 2*2=4 3*2=6
1*3=3 2*3=6 3*3=9
1*4=4 2*4=8 3*4=12
1*5=5 2*5=10 3*5=15
1*6=6 2*6=12 3*6=18
1*7=7 2*7=14 3*7=21
1*8=8 2*8=16 3*8=24
1*9=9 2*9=18 3*9=27
use "c[i]" or "c[i]c[i+1]" % a[1]
if (c[i]%a[1]==0) then b[i] must be c[i]/a[1]
else if (c[i]c[i+1]%a[i]==0) b[i] must be c[i]c[i+1]/a[1]
they won't be true at the same time.
Because,when x*y>=10,x*y%10 10
so,c[i]=c[i]c[i+1]%10=10 even though c[i]c[i+1]%a[1]==0
```
【代码】
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int n,m,l;
char s[N+10];
int a[N+10],b[N+10],c[N+10];
bool test(int a1){
a[1] = a1;
//get string b
int cur = 1;
for (int i = 1;i <= m;i++){//for b's each position
if (cur>l) return false;
//break limit,unaccpted
if (c[cur]%a1==0){//only one position
b[i] = c[cur]/a1;//get b[i]
cur++;
}else if (cur+1<=l && (c[cur]*10+c[cur+1])%a1==0){//twoposition
b[i] = (c[cur]*10+c[cur+1])/a1;//get b[i]
if (b[i]>=10) return false;
cur+=2;
}else return false;//no accpted
}
//use b and c,get string a
for (int i = 2;i <= n;i++){
if (cur>l) return false;
if (c[cur]%b[1]==0){
a[i] = c[cur]/b[1];
cur++;
}else if (cur+1<=l && (c[cur]*10+c[cur+1])%b[1]==0) {
a[i] = (c[cur]*10+c[cur+1])/b[1];
if (a[i]>=10) return false;
cur+=2;
}else return false;
//got a[i]
//use a[i] to check c[cur],c[cur+1]....
for (int j = 2;j <= m;j++){
if (cur>l) return false;
int _ju = a[i]*b[j];
//if (a1==2) printf("%d\n",_ju);
if (_ju<=9 && c[cur]==_ju){
cur++;
}else if ( _ju>=10 && c[cur]==(_ju/10) && cur+1<=l &&
c[cur+1]==(_ju%10) ) cur+=2;
else return false;
}
}
//
if (cur!=l+1) return false;
return true;
}
int main(){
// freopen("rush.txt","r",stdin);
int T;
scanf("%d",&T);
while (T--){
scanf("%d%d",&n,&m);
scanf("%s",s+1);
l = strlen(s+1);
for (int i = 1;i <= l;i++) c[i] = s[i]-'0';
int ok = 0;
for (int i = 1;i <= 9;i++)
if (test(i)){
ok = 1;
break;
}
if (ok==0){
puts("Impossible");
}else{
for (int i = 1;i <= n;i++) printf("%d",a[i]);printf(" ");
for (int i = 1;i <= m;i++) printf("%d",b[i]);
puts("");
}
}
return 0;
}