$Description$
$Sol$
进制转化+高精度除法
$over$
$Code$
#include<bits/stdc++.h> #define il inline #define Rg register #define go(i,a,b) for(Rg int i=a;i<=b;++i) #define yes(i,a,b) for(Rg int i=a;i>=b;--i) #define mem(a,b) memset(a,b,sizeof(a)) #define ll long long #define db double using namespace std; il int read() { Rg int x=0,y=1;char c=getchar(); while(c<‘0‘||c>‘9‘){if(c==‘-‘)y=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=(x<<1)+(x<<3)+c-‘0‘;c=getchar();} return x*y; } const int N=1010; int T,n,a,b,ct,d[N],as[N]; int main() { T=read(); while(T--) { ct=0;a=read(),b=read(); string s;cin>>s;n=s.length(); go(i,1,n) { char c=s[i-1]; if(c>=‘0‘ && c<=‘9‘)d[n-i+1]=c-‘0‘; if(c>=‘A‘ && c<=‘Z‘)d[n-i+1]=c-‘A‘+10; if(c>=‘a‘ && c<=‘z‘)d[n-i+1]=c-‘a‘+36; } while(n) { yes(i,n,2){d[i-1]+=d[i]%b*a;d[i]/=b;} as[++ct]=d[1]%b;d[1]/=b; while(n && !d[n])n--; } printf("%d ",a);cout<<s<<endl;printf("%d ",b); yes(i,ct,1) { Rg int c=as[i]; if(c>=0 && c<=9)printf("%d",c); if(c>=10 && c<=35)printf("%c",(char)(c-10+‘A‘)); if(c>=36 && c<=61)printf("%c",(char)(c-36+‘a‘)); } printf("\n\n"); } return 0; }