ZJU-PAT 1010. Radix (25)

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:
2
Sample Input 2:
1 ab 1 2
Sample Output 2:
Impossible
[注意]
1.数据类型选择:输入10位数结合进制可得应选long long int,在某些情况下radix较大故也应该设置为long long int
2.算法选择:简单顺序遍历超时,根据数值随着进制单调增加考虑使用二分查找(下界选max(s[i])+1, 上界选max(s[tag]-->longint,下界))例子:11 b 1 10
3.算法优化剪枝:(1)当两个数均为1,输出2;(2)当两数相等但不为1时,输出radix (3)计算某进制对应的数时累计和超过num1即可返回结果。
#pragma warning (disable:4786)
#include
#include
#include
#include
#include
#include
using namespace std;

string a[3];
int tag,radix;

long long int tolongint(string s,int dig)
{
    long long int ans=0;
    long long int d=1;
    for(int i=s.size()-1;i>=0;i--)
    {
        int num;
        if(s[i]>='0' && s[i]<='9') num=s[i]-'0';
        else num=s[i]-'a'+10;

        ans=ans+num*d;
        d*=dig;
    }
    return ans;
}

int Maxnum(string s)
{
    int d=-1;
    for(int i=0;i='0' && s[i]<='9') num=s[i]-'0';
        else num=s[i]-'a'+10;

        if(num>d) d=num;
    }
    return d+1;
}

int main()
{
    while(cin>>a[1]>>a[2]>>tag>>radix)
    {
        if(a[1]=="1" && a[2]=="1")
        {
            cout<<"2"<num1) break;
        }

        if(flag) cout<

 
  
#pragma warning (disable:4786)
#include
#include
#include
#include
#include
#include
using namespace std;

string a[3];
long long int  tag,radix;
long long int num1;

long long int tolongint(string s,long long int  dig)
{
    long long int ans=0;
    long long int d=1;
    for(int i=s.size()-1;i>=0;i--)
    {
        int num;
        if(s[i]>='0' && s[i]<='9') num=s[i]-'0';
        else num=s[i]-'a'+10;

        ans=ans+num*d;
        d*=dig;
    }
    return ans;
}

int cmp(string s,long long int  dig)
{
    long long int ans=0;
    long long int d=1;
    for(int i=s.size()-1;i>=0;i--)
    {
        int num;
        if(s[i]>='0' && s[i]<='9') num=s[i]-'0';
        else num=s[i]-'a'+10;

        ans=ans+num*d;
        if(ans>num1) return 1;
        d*=dig;
    }
    if(ans==num1) return 0;
    return -1;
}

int Maxnum(string s)
{
    int d=-1;
    for(int i=0;i='0' && s[i]<='9') num=s[i]-'0';
        else num=s[i]-'a'+10;

        if(num>d) d=num;
    }
    return d+1;
}

long long int Max(long long int a,long long int b)
{
    return a>b?a:b;
}

long long int BinarySearch(int cur)
{
    long long int L=Maxnum(a[cur]);
    long long int R=Max(L,num1),M;

    while(L<=R)
    {
        M=(L+R)/2;
        int res=cmp(a[cur],M);

        if(res==0) return M;
        else if(res==1) R=M-1;
        else L=M+1;
    }
    return -1;
}

int main()
{
    while(cin>>a[1]>>a[2]>>tag>>radix)
    {
        if(a[1]=="1" && a[2]=="1")
        {
            cout<<"2"<
ZJU-PAT 1010. Radix (25)_第1张图片

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