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.
给出一对正整数,比如说6和110.等式6=110是正确的吗。如果6是十进制数而110为二进制数,那么答案是yes。
现在给出任意一对正整数N1和N2,给出一个数的基数,你需要找到另一个数的基数
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.
每个测试样例有一行,包含4个正整数如图所示。
N1和N2不超过10位。每一位上的数都比基数小,并且选自集合{0-9,a-z} 0-9代表十进制数0-9,a-z代表十进制数10-35.如果tag是1那么radix就是N1的基数,如果tag是2那么radix就是N2的基数
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.
输出使N1=N2成立的另一个数的基数。如果这个等式不可能成立,输出Impossible。如果解决方案有多种,输出基数尽可能小的那个。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
LL Map[256];
LL inf=(1LL<<63)-1;//long long的最大值2^63-1,注意加括号
void init(){
for(char c='0';c<='9';c++){
Map[c]=c-'0';
}
for(char c='a';c<='z';c++){
Map[c]=c-'a'+10;//将'a-z'映射到10-35
}
}
LL convertNum10(char a[],LL radix,LL t){//将a转换为十进制,t为上界
LL ans=0;
int len=strlen(a);
for(int i=0;i<len;i++){
ans=ans*radix+Map[a[i]];
if(ans<0||ans>t) return -1;//判断溢出
}
return ans;
}
int cmp(char N2[],LL radix,LL t){//N2的十进制与t比较
int len=strlen(N2);
LL num=convertNum10(N2,radix,t);//将N2转换为十进制
if(num<0) return 1;//溢出,肯定是N2>t
if(t>num) return -1;//t较大,返回-1
else if(t==num) return 0;//相等,返回0
else return 1;//num较大,返回1
}
LL binarySearch(char N2[],LL left,LL right,LL t){
LL mid;
while(left<=right){
mid=(left+right)/2;
int flag=cmp(N2,mid,t);//判断N2转换为十进制后与t比较
if(flag==0) return mid;//找到解,返回mid
else if(flag == -1) left=mid+1;
else right=mid-1;
}
return -1;//解不存在
}
int findLargestDigit(char N2[]){//求最大的数位
int ans=-1,len=strlen(N2);
for(int i=0;i<len;i++){
if(Map[N2[i]]>ans){
ans=Map[N2[i]];
}
}
return ans+1;//最大的数位是ans,说明进制数的底线是ans+1
}
char N1[20],N2[20],temp[20];
int tag,radix;
int main(){
init();
scanf("%s %s %d %d",N1,N2,&tag,&radix);
if(tag==2){//让给出进制的做N1
strcpy(temp,N1);
strcpy(N1,N2);
strcpy(N2,temp);
}
LL t=convertNum10(N1,radix,inf);//将N1从radix进制转换成十进制
LL low=findLargestDigit(N2);//找到N2进制下界
LL high=max(low,t)+1;//找到N2进制上界
LL ans = binarySearch(N2,low,high,t);//二分查找N2进制
if(ans==-1) printf("Impossible\n");
else printf("%lld\n",ans);
return 0;
}
考虑到对一个确定的数字来说,它的进制越大,则将该数字串转换为十进制的结果也就越大,因此可以使用二分法。二分N2的金辉,将N2从该进制转换为十进制与N1的十进制比较。如果大于N1的十进制,说明N2当前进制太大,应往左子区间继续二分;如果小于N2的十进制,说明N2的当前进制太小,应往右子区间继续二分。
注意:使用遍历进制的暴力枚举会超时