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hdu----(5050)Divided Land(二进制求最大公约数)

2024-07-24 17:56:53   214人阅读

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
310 100100 11010010 1100
 

 

Sample Output
Case #1: 10Case #2: 10Case #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  else101   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 }
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hdu----(5050)Divided Land(二进制求最大公约数)


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