PAT甲级 1010 Radix

PAT甲级 1010 Radix

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.

Now for any pair of positive integers N1 and N​2​​ , your task is to find the radix of one number while that of the other is given.

Input Specification:

Each input file contains one test case. Each case occupies a line which contains 4 positive integers:

$N_1{N}_{2}tagradix$

Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2.

Output Specification:

For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix.

Sample Input 1:

6 110 1 10

Sample Output 1:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

这道题的陷阱就是很多人以为进制固定在2-36进制，其实不然，要用二分查找符合条件的进制，下限即为字符串最大的字符+1(注意下限最小为2)，上限就是确定进制的字符串的十进制，注意本题卡int，故须判断一下上界是否溢出。我写得尽可能简洁易懂，供大家参考，AC代码如下：

#include
using namespace std;
typedef long long ll;
ll ans1=0,ans2=0;
int n,k;
string s1,s2;
ll p(ll k,int t){//手写pow函数，防止其越界
    ll s=1;
    for(int i=0;i<t;i++){
        s*=k;
    }
    return s;
}

ll sum(string s,ll k){//k进制转十进制
    int len=s.length();
    ll ans=0;
    for(int i=0;i<len;i++){
        if(isdigit(s[i])) ans+=p(k,len-i-1)*(s[i]-'0');
        else ans+=p(k,len-i-1)*(s[i]-'a'+10);
    }
    return ans;
}

ll maxnum(string s){//找二分查找的下限
    int len=s.length();
    int ans=0;
    for(int i=0;i<len;i++){
        if(isdigit(s[i])) ans=max(ans,(s[i]-'0'));
        else ans=max(ans,s[i]-'a'+10);
    }
    return ll(ans+1);
}

ll bs(string s,ll a){//二分查找
    ll l=maxnum(s),r=max(a,l);
    if(l<2) l=2;//稳定下界，但是后来试了一下，不加也是对的，看来数据还没那么坑
    while(l<=r){
        ll mid=(l+r)/2;
        ll ans=sum(s,mid);
        if(ans==a) return mid;
        else if(ans>a || ans<0) r=mid-1;//ans<0很重要，检查上界是否溢出
        else l=mid+1;
    }
    return -1;
}

void judge(){
    ans1=sum(s1,k);
    ll p=bs(s2,ans1);
    if(p!=-1){cout<<p;return;}
    else{cout<<"Impossible";return;}
}

int main(){
    cin>>s1>>s2>>n>>k;
    if(n==2) swap(s1,s2);
    judge();
    return 0;
}