In addition to high-precision

Portal: http: //ybt.ssoier.cn: 8088 / problem_show.php pid = 1308?

 

Description [title]

High-precision divided by the high refinement, and their quotient and remainder.

[Enter]

Two input 300 is less than the positive integer.

[Output]

Output quotient and remainder.

[Sample input]

1231312318457577687897987642324567864324567876543245671425346756786867867867
1231312318767141738178325678412414124141425346756786867867867

[Sample Output]

999999999748590
179780909068307566598992807564736854549985603543237528310337 



dividend if it is 9, the divisor is 6, then adding the two high-precision subtraction divisor, each time adding provider 100 (the number of 0 bits of the dividend divisor is subtracted median -1) to. 

Examples: 
9745/3

the first time	9745	300	Save 32, I 145	Business 32 * 100
the second time	145	30	Minus 4 times, more than 25	List 32 * 100 * 10 + 4
the third time	25	3	Minus eight times, more than 1	List 32 * 100 + 4 * 10 + 8 * 1

code：

#include<iostream>
#include<cstring>
#define N 310
using namespace std;
string a,b;
int as[310],bs[310],ans[310],lena,lenb;
void jian(){
    for(int i=0;i<lena;i++)as[i]-=bs[i];
    for(int i=0;i<lena;i++)
    {
        if(as[i]<0){
            as[i]+=10;
            as[i+1]--;
        }
    }
    while(as[lena-1]==0&&lena>0)lena--;
}
bool dx(){
    if(lena>lenb)return true;
    if(lena<lenb)return false;
    for(int i=lena-1;i>=0;i--){
        if(as[i]>bs[i])return true;
        if(as[i]<bs[i])return false;
    }
    return true;
}
void you ( int k)
{
    if(k==0)k=1;
    if(k!=1)
    {
        for(int i=lenb-1;i>=0;i--)
        {   
            bs [i + k- 1 ] = bs [i];
            bs [i] = 0 ;
        }
        lenb+=k-1;
    }
    while(dx()==true){
        jian ();
        ans[k-1]++;
        int ls=k-1;
        for(int i=k-1;i<=ls;i++)
        {
            if(ans[i]>9){
                years [i + 1 ] ++ ;
                ans[i]-=10;
                if(ans[i+1]>9)ls++;
            }
        }
    }
    if(k!=1)
    {
        for(int i=k-1;i<lenb;i++)
        {
            bs [in -k + 1 ] = bs [i];
            bs [i] = 0 ;
        }
        lenb-=k-1;
    }
}
int main () {
    cin>>a>>b;
    if(a==b){cout<<1<<endl<<0<<endl;return 0;}
    lena=a.size();
    lenb=b.size();
    for(int i=0;i<lena;i++)as[i]=a[lena-i-1]-'0';
    for(int i=0;i<lenb;i++)bs[i]=b[lenb-i-1]-'0';
    memset(ans,0,sizeof(ans));
    int k=lena-lenb;
    if(k<0){cout<<0<<endl<<a<<endl;return 0;}
    while(k>=0){
        if(k==0&&dx()==false)break;
        you (k);
        k=lena-lenb;
    }
    int h1=301,h2=301;
    while(ans[h1]==0&&h1>0)h1--;
    while(as[h2]==0&&h2>0)h2--;
    for(int i=h1;i>=0;i--)cout<<ans[i];
    cout<<endl;
    for(int i=h2;i>=0;i--)cout<<as[i];
    cout<<endl;
}

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