toj3860

本题略坑，给出的长度可能小于上面给出的进制的个数

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;char a[50];int b[50];int c[50];int main(){    int n;    while(scanf("%d",&n)==1){        for(int i=0;i<n;i++){            cin>>a[i];//将字母转化为数字            if('0'<=a[i]&&a[i]<='9')                b[i]=a[i]-'0';            else{                b[i]=a[i]-'A'+10;            }        }        string s;        cin>>s;        int len=s.size();        for(int i=0;i<len;i++){            if('0'<=s[i]&&s[i]<='9')                c[i]=s[i]-'0';            else{                c[i]=s[i]-'A'+10;            }        }        ll ans=0;        ll base1=1;//考虑两者不相等的情况        for(int i=n-1,j=len-1;i>=0&&j>=0;i--,j--){            ans=ans+base1*c[j];            base1=base1*b[i];         }        cout<<ans<<endl;    }}