Divided Land

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 56    Accepted Submission(s): 27 

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)

OutputFor 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

Source

2014 ACM/ICPC Asia Regional Shanghai Online

#include <stdio.h>
#include <string.h>
#define MAXN 1000
struct BigNumber
{
    int len;
    int v[MAXN];
};
bool isSmaller(BigNumber n1,BigNumber n2)
{
    if(n1.len<n2.len)
        return 1;
    if(n1.len>n2.len)
        return 0;
    for(int i=n1.len-1; i>=0; i--)
    {
        if(n1.v[i]<n2.v[i])
            return 1;
        if(n1.v[i]>n2.v[i])
            return 0;
    }
    return 0;
}
BigNumber minus(BigNumber n1,BigNumber n2)
{
    BigNumber ret;
    int borrow,i,temp;
    ret=n1;
    for(borrow=0,i=0; i<n2.len; i++)
    {
        temp=ret.v[i]-borrow-n2.v[i];
        if(temp>=0)
        {
            borrow=0;
            ret.v[i]=temp;
        }
        else
        {
            borrow=1;
            ret.v[i]=temp+2;
        }
    }
    for(; i<n1.len; i++)
    {
        temp=ret.v[i]-borrow;
        if(temp>=0)
        {
            borrow=0;
            ret.v[i]=temp;
        }
        else
        {
            borrow=1;
            ret.v[i]=temp+2;
        }
    }
    while(ret.len>=1 && !ret.v[ret.len-1])
        ret.len--;
    return ret;
}
BigNumber div2(BigNumber n)
{
    BigNumber ret;
    ret.len=n.len-1;
    for(int i=0; i<ret.len; i++)
        ret.v[i]=n.v[i+1];
    return ret;
}
void gcd(BigNumber n1,BigNumber n2)
{
    long b=0,i;
    while(n1.len && n2.len)
    {
        if(n1.v[0])
        {
            if(n2.v[0])
            {
                if(isSmaller(n1,n2))
                    n2=minus(n2,n1);
                else
                    n1=minus(n1,n2);
            }
            else
                n2=div2(n2);
        }
        else
        {
            if(n2.v[0])
                n1=div2(n1);
            else
            {
                n1=div2(n1);
                n2=div2(n2);
                b++;
            }
        }
    }
    if(n2.len)
        for(i=n2.len-1; i>=0; i--)
            printf("%d",n2.v[i]);
    else
        for(i=n1.len-1; i>=0; i--)
            printf("%d",n1.v[i]);
    while(b--)
        printf("0");
    printf("\n");
}
int main()
{
    int cases,le,i;
    BigNumber n1,n2;
    char str1[MAXN],str2[MAXN];
    scanf("%d",&cases);
    int num;
    for(num=1;num<=cases;num++)
    {
        scanf("%s%s",str1,str2);
        le=strlen(str1);
        n1.len=le;
        for(i=0; i<le; i++)
            n1.v[i]=str1[le-1-i]-‘0‘;
        le=strlen(str2);
        n2.len=le;
        for(i=0; i<le; i++)
            n2.v[i]=str2[le-1-i]-‘0‘;
            printf("Case #%d: ",num);
        gcd(n1,n2);
    }
    return 0;
}