Divided Land

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 123    Accepted Submission(s): 64

Problem Description

It’s time to fight the local despots and redistribute the land. There
is a rectangular piece of land granted from the government, whose
length and width are both in binary form. As the mayor, you must segment
the land into multiple squares of equal size for the villagers. What
are required is there must be no any waste and each single segmented
square land has as large area as possible. The width of the segmented
square land is also binary.

Input

The first line of the input is T (1 ≤ T ≤ 100), which stands for the number of test cases you need to solve.

Each case contains two binary number represents the length L and the width W of given land. (0 < L, W ≤ 2¹⁰⁰⁰)

Output

For each test case, print a line “Case #t: ”(without quotes, t means
the index of the test case) at the beginning. Then one number means the
largest width of land that can be divided from input data. And it will
be show in binary. Do not have any useless number or space.

Sample Input

3
10 100
100 110
10010 1100

Sample Output

Case #1: 10
Case #2: 10
Case #3: 110

二进制求最大公约数：

代码：

  1 #include <stdio.h>
  2 #include <string.h>
  3 #define MAXN 1000
  4 struct BigNumber{
  5    int len;
  6    int v[MAXN];
  7 };
  8 bool isSmaller(BigNumber n1,BigNumber n2)
  9 {
 10  if(n1.len<n2.len)
 11   return 1;
 12  if(n1.len>n2.len)
 13   return 0;
 14  for(int i=n1.len-1;i>=0;i--)
 15  {
 16   if(n1.v[i]<n2.v[i])
 17    return 1;
 18   if(n1.v[i]>n2.v[i])
 19    return 0;
 20  }
 21  return 0;
 22 }
 23 BigNumber minus(BigNumber n1,BigNumber n2)
 24 {
 25  BigNumber ret;
 26  int borrow,i,temp;
 27  ret=n1;
 28  for(borrow=0,i=0;i<n2.len;i++)
 29  {
 30   temp=ret.v[i]-borrow-n2.v[i];
 31   if(temp>=0)
 32   {
 33    borrow=0;
 34    ret.v[i]=temp;
 35   }
 36   else
 37   {
 38    borrow=1;
 39    ret.v[i]=temp+2;
 40   }
 41  }
 42  for(;i<n1.len;i++)
 43  {
 44   temp=ret.v[i]-borrow;
 45   if(temp>=0)
 46   {
 47    borrow=0;
 48    ret.v[i]=temp;
 49   }
 50   else
 51   {
 52    borrow=1;
 53    ret.v[i]=temp+2;
 54   }
 55  }
 56  while(ret.len>=1 && !ret.v[ret.len-1])
 57   ret.len--;
 58  return ret;
 59 }
 60 BigNumber div2(BigNumber n)
 61 {
 62  BigNumber ret;
 63  ret.len=n.len-1;
 64  for(int i=0;i<ret.len;i++)
 65   ret.v[i]=n.v[i+1];
 66  return ret;
 67 }
 68 void gcd(BigNumber n1,BigNumber n2)
 69 {
 70  long b=0,i;
 71  while(n1.len && n2.len)
 72  {
 73   if(n1.v[0])
 74   {
 75    if(n2.v[0])
 76    {
 77     if(isSmaller(n1,n2))
 78      n2=minus(n2,n1);
 79     else
 80      n1=minus(n1,n2);
 81    }
 82    else
 83     n2=div2(n2);
 84   }
 85   else
 86   {
 87    if(n2.v[0])
 88     n1=div2(n1);
 89    else
 90    {
 91     n1=div2(n1);
 92     n2=div2(n2);
 93     b++;
 94    }
 95   }
 96  }
 97  if(n2.len)
 98   for(i=n2.len-1;i>=0;i--)
 99    printf("%d",n2.v[i]);
100  else
101   for(i=n1.len-1;i>=0;i--)
102    printf("%d",n1.v[i]);
103  while(b--)
104   printf("0");
105  printf("\n");
106 }
107 int main()
108 {
109  int cases,le,i;
110  BigNumber n1,n2;
111  char str1[MAXN],str2[MAXN];
112  scanf("%d",&cases);
113  for(int w=1;w<=cases;w++)
114  {
115   scanf("%s%s",str1,str2);
116   le=strlen(str1);
117   n1.len=le;
118   for(i=0;i<le;i++)
119    n1.v[i]=str1[le-1-i]-‘0‘;
120   le=strlen(str2);
121   n2.len=le;
122   for(i=0;i<le;i++)
123    n2.v[i]=str2[le-1-i]-‘0‘;
124    printf("Case #%d: ",w);
125   gcd(n1,n2);
126  }
127  return 0;
128 }