PAT甲级-1010. Radix (25)进制

柔光的暖阳◎ 2022-05-31 09:52 177阅读 0赞

# 1010. Radix (25) #

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 N2, 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:  
N1 N2 tag radix  
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:**

**Sample Input 2:**

1 ab 1 2

**Sample Output 2:**

Impossible

给出两个数N1和N2，Tag指出所给的Radix是第几个数的进制，计算出一个进制使得N1=N2。

我的思路是先将给定进制的数转换成十进制sum，记录下未定数中各个位上最大的一位数mmax，然后根据“一个数的各个位上最大的数小于其进制”在\[mmax+1，sum+1\]范围内二分遍历可能使得两数相等的进制。

(；´д｀)ゞ这里如果不二分的话，for循环在测试点7上限需要开到1e10，会直接超时爆掉…

#include<bits/stdc++.h>
    using namespace std;
    #define INF 0xfffffff
    #define MAXN 1010
    int a[MAXN],b[MAXN],mmax;
    long long sum,res,ans;
    void change10(string x,int r)//转成10进制
    {
        memset(a,0,sizeof(a));
        int len=x.length();
        int cnt=len-1;
        sum=0;
        for(int i=0; i<len; ++i)
        {
            if(x[i]>='0'&&x[i]<='9')
                a[i]=int(x[i]-'0');
            else if(x[i]>='a'&&x[i]<='z')
                a[i]=int(x[i]-'a'+10);
            sum+=(pow(r,cnt)*a[i]);
            --cnt;
        }
    }
    bool tryR(string x)//依次尝试各种进制
    {
        memset(b,0,sizeof(b));
        int len=x.length();
        mmax=0;
        for(int i=0; i<len; ++i)
        {
            if(x[i]>='0'&&x[i]<='9')
                b[i]=int(x[i]-'0'),mmax=max(mmax,b[i]);
            else if(x[i]>='a'&&x[i]<='z')
                b[i]=int(x[i]-'a'+10),mmax=max(mmax,b[i]);
        }
        long long low=mmax+1,high=sum+1;//上限至少要开到10000000000;
        long long mid;
    A:
        while(low<=high)//二分处理
        {
            res=0;
            mid=(low+high)/2;
            int cnt=len-1;
            for(int i=0; i<len; ++i)
            {
                res+=(pow(mid,cnt)*b[i]),--cnt;
                if(res>sum||res<0)//越界
                {
                    high=mid-1;
                    goto A;
                }
            }
            if(res<sum) low=mid+1;
            else if(res>sum) high=mid-1;
            else if(res==sum)//&&mid>=mmax
            {
                ans=mid;
                return true;
            }
        }
        return false;
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("F:/cb/read.txt","r",stdin);
    //freopen("F:/cb/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        string n,m;
        int t,r;
        cin>>n>>m>>t>>r;
        bool flag=false;
        if(t==1)
            change10(n,r),flag=tryR(m);
        else if(t==2)
            change10(m,r),flag=tryR(n);
        if(flag) cout<<ans<<endl;
        else cout<<"Impossible"<<endl;
    }