Decimal integer conversion

Decimal integer conversion

时间限制：1000 ms  |  内存限制：65535 KB

难度：2

描述

XiaoMing likes mathematics, and he is just learning how to convert numbers between different bases , but he keeps making errors since he is only 6 years old. Whenever XiaoMing converts a number to a new base and writes down the result, he always writes one of the digits wrong. For example , if he converts the number 14 into binary (i.e., base 2), the correct result should be "1110", but he might instead write down "0110" or "1111". XiaoMing never accidentally adds or deletes digits, so he might write down a number with a leading digit of " 0" if this is the digit she gets wrong. Given XiaoMing 's output when converting a number N into base 2 and base 3, please determine the correct original value of N (in base 10). (N<=10^10) You can assume N is at most 1 billion, and that there is a unique solution for N. 

输入

The first line of the input contains one integers T, which is the nember of test cases (1<=T<=8)
Each test case specifies:
* Line 1: The base-2 representation of N , with one digit written incorrectly.
* Line 2: The base-3 representation of N , with one digit written incorrectly.

输出

For each test case generate a single line containing a single integer , the correct value of N

样例输入

11010212

样例输出

14

来源

河南省第九届省赛

题目意思是：给你t组测试数据，每组数据，有两行，第一行是二进制，第二行是三进制。并且它们俩表示的十进制数字是同一个。

但是，每行都有一位数字是错误的，现在让你求出来原来的十进制数字。

#include <stdio.h>  #include <string.h>  #include <math.h>  int a[50];//二进制错误数组   int b[105];//二进制错误数组  char s1[50];//二进制字符串   char s2[50];//三进制字符串   int main()  {      int T,i,j,k;      scanf("%d",&T);      getchar();      while(T--)      {          gets(s1);          gets(s2);          int len1=strlen(s1);          int len2=strlen(s2);          memset(a,0,sizeof(a));          memset(b,0,sizeof(b));          for(i=0;i<len1;i++)          {              s1[i]=(s1[i]-'0'+1)%2+'0';//对进制取错位               for(j=0;j<len1;j++)                  a[i]+=(s1[j]-'0')*(int)pow(2,len1-1-j);              s1[i]=(s1[i]-'0'+1)%2+'0';//还原进制字符串，确保每次只有一位错误           }          for(i=0,k=0;i<len2;i++)          {              s2[i]=(s2[i]-'0'+1)%3+'0';//原理同二进制中的代码               for(j=0;j<len2;j++)                  b[k]+=(s2[j]-'0')*(int)pow(3,len2-1-j);              k++;              s2[i]=(s2[i]-'0'+1)%3+'0';//每次累加 1 取错位               for(j=0;j<len2;j++)                  b[k]+=(s2[j]-'0')*(int)pow(3,len2-1-j);              k++;              s2[i]=(s2[i]-'0'+1)%3+'0';//还原           }          for(i=0;i<len1;i++)              for(j=0;j<k;j++)                  if(a[i]==b[j])                  {                      printf("%d\n",a[i]);                      break;                  }      }      return 0;  }