标签：poj zoj 算法 c++ 编程
Multiply
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 5179 |
|
Accepted: 2773 |
Description
6*9 = 42" is not true for base 10, but is true for base 13. That is, 6(13) * 9(13) = 42(13) because 42(13) = 4 * 131 + 2 * 130 = 54(10).
You are to write a program which inputs three integers p, q, and r and determines the base B (2<=B<=16) for which p * q = r. If there are many candidates for B, output the smallest one. For example, let p = 11, q = 11, and r = 121. Then we have 11(3) * 11(3)
= 121(3) because 11(3) = 1 * 31 + 1 * 30 = 4(10) and 121(3) = 1 * 32 + 2 * 31 + 1 * 30 = 16(10). For another base such as 10, we also have 11(10) * 11(10) = 121(10). In this case, your program should output 3 which is the smallest base. If there is no candidate
for B, output 0.
Input
The input consists of T test cases. The number of test cases (T ) is given in the first line of the input file. Each test case consists of three integers p, q, and r in a line. All digits of p, q, and r are numeric digits and 1<=p,q, r<=1,000,000.
Output
Print exactly one line for each test case. The line should contain one integer which is the smallest base for which p * q = r. If there is no such base, your program should output 0.
Sample Input
3
6 9 42
11 11 121
2 2 2
Sample Output
13
3
0
题意：给你三个数字p，q，r，问在哪个最小的进制下p*q=r成立；
注意：如果在k进制下p，q，r的每位上的数字应该要小于k；
#include <iostream>
#include <string.h>
using namespace std;
int change(char *a,int k){
int len=strlen(a);
int ans=0;
for (int i=0;i<len;i++){
ans = ans*k + a[i]-'0';
}
return ans;
}
bool isBig(char *a,int k){
int len=strlen(a);
for (int i=0;i<len;i++){
if (a[i]-'0'>=k)
return false;
}
return true;
}
int main(){
char a[10],b[10],r[10];
int t,i;
cin>>t;
while (t--){
cin>>a>>b>>r;
for (i=2;i<=16;i++){
if (isBig(a,i)==true && isBig(b,i)==true && isBig(r,i)==true){
int aa = change(a,i);
int bb = change(b,i);
int rr = change(r,i);
long long ans = aa*bb;
long long ans2 = rr;
if (ans==ans2)
break;
}
}
if (i==17)
i=0;
cout<<i<<endl;
}
return 0;
}
poj 1331 Multiply
标签：poj zoj 算法 c++ 编程
原文地址：http://blog.csdn.net/codeforcer/article/details/42001369
