1010 Radix (25)（25 分）
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 10Sample Output 1:
2Sample Input 2:
1 ab 1 2Sample Output 2:
Impossible
题意分析：
给出两个正整数，给出其中一个正整数的基底，求另外一个正整数的基底，使得这两个正整数在各自的基底的十进制数相等
23分代码
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<map>
#define inf 0x3f3f3f3f
using namespace std;
int main()
{
int rag,radix;
char a[15];
char b[15];
int c[15];
int l;
while(cin>>a>>b>>rag>>radix)
{
int da=0,db;
if(rag==1)
{
int la=strlen(a);
int k=1;
for(int i=la-1;i>=0;i--)
{
if(a[i]>='a'&&a[i]<='z')
{
da+=(a[i]-'a'+10)*k;
}
else
da+=(a[i]-'0')*k;
k=k*radix;
}
l=strlen(b);
for(int i=l-1;i>=0;i--)
{
if(b[i]>='a'&&b[i]<='z')
{
c[i]=b[i]-'a'+10;
}
else
c[i]=b[i]-'0';
}
}
if(rag==2)
{
int lb=strlen(b);
int k=1;
for(int i=lb-1;i>=0;i--)
{
if(b[i]>='a'&&b[i]<='z')
{
da+=(b[i]-'a'+10)*k;
}
else
da+=(b[i]-'0')*k;
k=k*radix;
}
l=strlen(a);
for(int i=l-1;i>=0;i--)
{
if(a[i]>='a'&&a[i]<='z')
{
c[i]=a[i]-'a'+10;
}
else
c[i]=a[i]-'0';
}
}
int ma=0;
for(int i=0;i<l;i++)
{
if(c[i]>ma)
ma=c[i];
}
bool f=0;
for(int i=ma+1;i<=1000005;i++)
{
db=0;
int k=1;
for(int j=l-1;j>=0;j--)
{
db+=c[j]*k;
k=k*i;
}
if(db==da)
{
f=1;
cout<<i<<endl;
break;
}
}
if(f==0)
cout<<"Impossible"<<endl;
}
return 0;
}
AC代码：
要用long long，要用二分，否则有两个数据点会超时（第7和第18个数据点），二分时，left还好，right居然要这么大（否则第7个数据点卡死），而且，算出来的数<0也代表数太大，爆long long ，要ri=mid-1;而且，题目说有多个要挑出最小的那个。
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<map>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int main()
{
ll rag,radix;
char a[15];
char b[15];
int c[15];
ll l;
while(cin>>a>>b>>rag>>radix)
{
ll da=0,db;
if(rag==1)
{
ll la=strlen(a);
ll k=1;
for(ll i=la-1;i>=0;i--)
{
if(a[i]>='a'&&a[i]<='z')
{
da+=(a[i]-'a'+10)*k;
}
else
da+=(a[i]-'0')*k;
k=k*radix;
}
l=strlen(b);
for(ll i=l-1;i>=0;i--)
{
if(b[i]>='a'&&b[i]<='z')
{
c[i]=b[i]-'a'+10;
}
else
c[i]=b[i]-'0';
}
}
if(rag==2)
{
ll lb=strlen(b);
ll k=1;
for(ll i=lb-1;i>=0;i--)
{
if(b[i]>='a'&&b[i]<='z')
{
da+=(b[i]-'a'+10)*k;
}
else
da+=(b[i]-'0')*k;
k=k*radix;
}
l=strlen(a);
for(ll i=l-1;i>=0;i--)
{
if(a[i]>='a'&&a[i]<='z')
{
c[i]=a[i]-'a'+10;
}
else
c[i]=a[i]-'0';
}
}
ll ma=0;
for(ll i=0;i<l;i++)
{
if(c[i]>ma)
ma=c[i];
}
ll le=ma+1,ri=10000005;
bool f=0;
ll mm=9999999999;
while(le<=ri)
{
ll mid=(ri+le)/2;
db=0;
ll k=1;
for(ll j=l-1;j>=0;j--)
{
db+=c[j]*k;
k=mid*k;
}
if(db==da)
{
if(mid<mm)//题目有多个要挑出最小的那个
mm=mid;
ri=mid-1;
f=1;
}
else if(db>da||db<0)//算出来的数<0也代表数太大，爆long long ，要ri=mid-1;
{
ri=mid-1;
}
else if(db<da)
{
le=mid+1;
}
}
if(f==0)
cout<<"Impossible"<<endl;
else
{
cout<<mm<<endl;
}
}
return 0;
}
1010 Radix (25)（25 分）
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 10Sample Output 1:
2Sample Input 2:
1 ab 1 2Sample Output 2:
Impossible
题意分析：
给出两个正整数，给出其中一个正整数的基底，求另外一个正整数的基底，使得这两个正整数在各自的基底的十进制数相等
23分代码
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<map>
#define inf 0x3f3f3f3f
using namespace std;
int main()
{
int rag,radix;
char a[15];
char b[15];
int c[15];
int l;
while(cin>>a>>b>>rag>>radix)
{
int da=0,db;
if(rag==1)
{
int la=strlen(a);
int k=1;
for(int i=la-1;i>=0;i--)
{
if(a[i]>='a'&&a[i]<='z')
{
da+=(a[i]-'a'+10)*k;
}
else
da+=(a[i]-'0')*k;
k=k*radix;
}
l=strlen(b);
for(int i=l-1;i>=0;i--)
{
if(b[i]>='a'&&b[i]<='z')
{
c[i]=b[i]-'a'+10;
}
else
c[i]=b[i]-'0';
}
}
if(rag==2)
{
int lb=strlen(b);
int k=1;
for(int i=lb-1;i>=0;i--)
{
if(b[i]>='a'&&b[i]<='z')
{
da+=(b[i]-'a'+10)*k;
}
else
da+=(b[i]-'0')*k;
k=k*radix;
}
l=strlen(a);
for(int i=l-1;i>=0;i--)
{
if(a[i]>='a'&&a[i]<='z')
{
c[i]=a[i]-'a'+10;
}
else
c[i]=a[i]-'0';
}
}
int ma=0;
for(int i=0;i<l;i++)
{
if(c[i]>ma)
ma=c[i];
}
bool f=0;
for(int i=ma+1;i<=1000005;i++)
{
db=0;
int k=1;
for(int j=l-1;j>=0;j--)
{
db+=c[j]*k;
k=k*i;
}
if(db==da)
{
f=1;
cout<<i<<endl;
break;
}
}
if(f==0)
cout<<"Impossible"<<endl;
}
return 0;
}
AC代码：
要用long long，要用二分，否则有两个数据点会超时（第7和第18个数据点），二分时，left还好，right居然要这么大（否则第7个数据点卡死），而且，算出来的数<0也代表数太大，爆long long ，要ri=mid-1;而且，题目说有多个要挑出最小的那个。
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#include<map>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int main()
{
ll rag,radix;
char a[15];
char b[15];
int c[15];
ll l;
while(cin>>a>>b>>rag>>radix)
{
ll da=0,db;
if(rag==1)
{
ll la=strlen(a);
ll k=1;
for(ll i=la-1;i>=0;i--)
{
if(a[i]>='a'&&a[i]<='z')
{
da+=(a[i]-'a'+10)*k;
}
else
da+=(a[i]-'0')*k;
k=k*radix;
}
l=strlen(b);
for(ll i=l-1;i>=0;i--)
{
if(b[i]>='a'&&b[i]<='z')
{
c[i]=b[i]-'a'+10;
}
else
c[i]=b[i]-'0';
}
}
if(rag==2)
{
ll lb=strlen(b);
ll k=1;
for(ll i=lb-1;i>=0;i--)
{
if(b[i]>='a'&&b[i]<='z')
{
da+=(b[i]-'a'+10)*k;
}
else
da+=(b[i]-'0')*k;
k=k*radix;
}
l=strlen(a);
for(ll i=l-1;i>=0;i--)
{
if(a[i]>='a'&&a[i]<='z')
{
c[i]=a[i]-'a'+10;
}
else
c[i]=a[i]-'0';
}
}
ll ma=0;
for(ll i=0;i<l;i++)
{
if(c[i]>ma)
ma=c[i];
}
ll le=ma+1,ri=10000005;
bool f=0;
ll mm=9999999999;
while(le<=ri)
{
ll mid=(ri+le)/2;
db=0;
ll k=1;
for(ll j=l-1;j>=0;j--)
{
db+=c[j]*k;
k=mid*k;
}
if(db==da)
{
if(mid<mm)//题目有多个要挑出最小的那个
mm=mid;
ri=mid-1;
f=1;
}
else if(db>da||db<0)//算出来的数<0也代表数太大，爆long long ，要ri=mid-1;
{
ri=mid-1;
}
else if(db<da)
{
le=mid+1;
}
}
if(f==0)
cout<<"Impossible"<<endl;
else
{
cout<<mm<<endl;
}
}
return 0;
}
