高精度乘以高精度

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5;
char A[N],B[N];
int a[N],b[N],c[N],na,nb;
void change(){
    for(int i=0; i<na; i++){
        a[na-i] = A[i] - '0';
    }
    for(int i=0; i<nb; i++){
        b[nb-i] = B[i] - '0';
    }
}
int main(){
    scanf("%s%s",A,B);
    na = strlen(A);
    nb = strlen(B);
    change();
    for(int i=1; i<=na; i++){
        for(int j=1; j<=nb; j++){
            c[i+j-1] += a[i]*b[j];
            c[i+j] += c[i+j-1] / 10;
            c[i+j-1] %= 10;
        }
    }
    int nc = na+nb;
    while(nc > 1 && !c[nc]) nc--;
    for(int i=nc; i>=1; i--){
        printf("%d",c[i]);
    }
    return 0;
}