nyoj 1277 Decimal integer conversion（水题）

Decimal integer conversion

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

1

1010

212

样例输出

14

来源

河南省第九届省赛

ps：大水题一个，暴力枚举就好了，但是我写的代码不堪直视啊。。。

先看我照大神思路写的代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
int a[100],b[100];

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
char s1[100],s2[100];
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
scanf("%s %s",s1,s2);
int len1=strlen(s1);
int len2=strlen(s2);
for(int i=0; i<len1; ++i)
{
s1[i]=(s1[i]-'0'+1)%2+'0';//取错位
for(int j=len1-1; ~j; --j)
a[i]+=(s1[j]-'0')*(1<<(len1-1-j));
s1[i]=(s1[i]-'0'+1)%2+'0';//还原
}
int k=0;
for(int i=0; i<len2; ++i,++k)
{
s2[i]=(s2[i]-'0'+1)%3+'0';//每次累加 1 取错位
for(int j=len2-1; ~j; --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(int j=len2-1; ~j; --j)
b[k]+=(s2[j]-'0')*(int)pow(3,len2-1-j);
s2[i]=(s2[i]-'0'+1)%3+'0';//还原
}
int flag=0;
for(int i=0; i<len1; ++i)
{
for(int j=0; j<k; ++j)
if(a[i]==b[j])
{
printf("%d\n",a[i]);
flag=1;
break;
}
if(flag) break;
}
}
return 0;
}

本人代码（手动捂脸）：

#include<stdio.h>
#include<string.h>

int change1(int a[],int len)//二进制转换
{
int k=1,num=0;
for(int i=len-1; ~i; --i)
{
if(a[i]==1)
num+=k;
k<<=1;
}
return num;
}

int change2(int b[],int len)//三进制转换
{
int k=1,num=0;
for(int i=len-1; ~i; --i)
{
if(b[i]==1)
num+=k;
else if(b[i]==2)
num+=k*2;
k*=3;
}
return num;
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
char s1[100],s2[100];
int a[100],b[100];
int flag=0,ans,tot1,tot2;
scanf("%s %s",s1,s2);
int len1=strlen(s1);
int len2=strlen(s2);
for(int i=0; i<len1; ++i)
{
if(s1[i]=='0')
{
for(int j=0; j<len1; ++j)
if(i==j)
a[j]=1;
else
a[j]=s1[j]-'0';
tot1=change1(a,len1);
for(int l=0; l<len2; ++l)
{
if(s2[l]=='0')
{
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=1;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=2;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
}
else if(s2[l]=='1')
{
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=0;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=2;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
}
else
{
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=1;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=0;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
}
}
if(flag)
break;
}
else
{
for(int j=0; j<len1; ++j)
if(i==j)
a[j]=2;
else
a[j]=s1[j]-'0';
tot1=change1(a,len1);
for(int l=0; l<len2; ++l)
{
if(s2[l]=='0')
{
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=1;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=2;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
}
else if(s2[l]=='1')
{
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=0;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=2;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
}
else
{
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=1;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
for(int k=0; k<len2; ++k)
if(k==l)
b[k]=0;
else
b[k]=s2[k]-'0';
tot2=change2(b,len2);
if(tot1==tot2)
{
flag=1;
ans=tot1;
break;
}
}
}
if(flag)
break;
}
}
printf("%d\n",ans);
}
return 0;
}