总时间限制: 1000ms 内存限制: 65536kB 描述 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. 输入 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. 输出 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. 样例输入 3 6 9 42 11 11 121 2 2 2 样例输出 13 3 0
#include <iostream> #include<string.h> using namespace std; //http://bailian.openjudge.cn/exam/3658/3/ //b进制转10进制，一般都是用char数组存储的 int n,flag; char p[8],q[8],r[8]; int pp,qq,rr; int b2t(char x[],int b){ int res=0,len=strlen(x); for(int i=0;i<len;i++){ if(x[i]-'0'>=b){ return -1; } else{ res=res*b+(x[i]-'0'); } } return res; } int main(int argc, char *argv[]) { cin>>n; while(n--){ flag=0; cin>>p>>q>>r; for(int i=2;i<=16;i++){ pp=b2t(p,i); if(pp!=-1){ qq=b2t(q,i); if(qq!=-1){ rr=b2t(r,i); if(rr!=-1&&rr==pp*qq){ cout<<i<<endl; flag=1; break; } } } } if(flag==0){ cout<<"0"<<endl; } } return 0; }