题目链接:https://www.luogu.org/problemnew/show/P1015
题目大意:给一个进制为
的数
,若经过
次使得
为回文数,输出
,若
,则输出 Impossible!
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
string m;
int a[100+10], b[100+10];
bool check(string s){ //检查是否为回文数
int len = s.length();
for(int i = 0;i < len/2; i++){
if(s[i] != s[len - i - 1]) return false;
}
return true;
}
string add(string s){
memset(a, 0, sizeof a),memset(b, 0, sizeof b);
int len = s.length();
for(int i = 0;i < len;i++){
if(s[i] >='0' && s[i] <= '9') b[i] = a[len-i-1] = s[i] - '0';
else if(s[i]>='a' && s[i] <='f') b[i] = a[len-i-1] = s[i] - 'a' + 10;
else if(s[i]>='A' && s[i] <='F') b[i] = a[len-i-1] = s[i] - 'A' + 10;
}
for(int i = 0; i < len; i++){
a[i]+=b[i];a[i+1] += a[i]/n; a[i]%=n;
}
if(a[len]) len++;
string res="";
for(int i=0;i<len;i++){
if(a[i]>=10) res+=a[i]-10 +'A';
else res+=a[i]+'0';
}
return res;
}
int main()
{
cin>>n>>m;
int res = 0;
while(!check(m)){
m = add(m);
res++;
if(res>=30) {
cout<<"Impossible!"<<endl;
return 0;
}
}
cout<<"STEP="<<res<<endl;
return 0;
}