大数处理

<p style="color: rgb(51,51); font-family: Arial; font-size: 14px; line-height: 26px;">题意：给定一个长方形，将该长方形切分成几个面积相等且边最大的正方形，求这样的正方形的最大边长，难点在于长方形边长为二进制形式，输出的正方形边长也为二进制，另外该题数据量巨大2^1000。</p><p style="color: rgb(51,51); font-family: Arial; font-size: 14px; line-height: 26px;">解题思路：由于该题数据量巨大，所以要用到大数处理方法，不过不用担心，对于大数有专门的模板，可以说这道题是模板题，直接看代码吧。</p>

#include <iostream>
#include <cstdio>
#include <cstring>
#define MAXN 1000
using namespace std;

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,int j)
{
    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++;
            }
        }
    }
    printf("Case #%d: ",j);
    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);
    for(int j = 1; j <= cases; j++)
    {
        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';
            gcd(n1,n2,j);
    }
    return 0;
}