-其它- A+B|A-B|A*B Problem

当long long不够用时，就得用高精度了！

高精度加法

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char a1[500001],b1[500001];
int a[500001],b[500001],c[500001],x,lenc,la,lb,i;
int main(){
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    gets(a1);
    gets(b1);
    la=strlen(a1);
    lb=strlen(b1);
    for(i=0;i<=la-1;i++){
        a[la-i]=a1[i]-'0';
    }
    for(i=0;i<=lb-1;i++){
        b[lb-i]=b1[i]-'0';
    }
    lenc=1;
    x=0;
    while(lenc<=la||lenc<=lb){
        c[lenc]=a[lenc]+b[lenc]+x;
        x=c[lenc]/10;
        c[lenc]%=10;
        lenc++;
    }
    c[lenc]=x;
    if(c[lenc]==0) lenc--;
    for(i=lenc;i>=1;i--) cout<<int(c[i]);
    return 0;
}

高精度减法

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main(){
    int a[10001],b[10001],c[10001],la,lb,lc,i;
    char n[10001],n1[10001],n2[10001];
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    cin>>n1>>n2;
    if(strlen(n1)<strlen(n2)||(strlen(n1)==strlen(n2)&&strcmp(n1,n2)<0)){
        strcpy(n,n1);
        strcpy(n1,n2);
        strcpy(n2,n);
        cout<<"-";
    }
    la=strlen(n1);
    lb=strlen(n2);
    for(i=0;i<=la-1;i++){
        a[la-i]=int(n1[i]-'0');
    }
    for(i=0;i<=lb-1;i++){
        b[lb-i]=int(n2[i]-'0');
    }
    i=1;
    while(i<=la||i<=lb){
        if(a[i]<b[i]){
            a[i]+=10;
            a[i+1]--;
        }
        c[i]=a[i]-b[i];
        i++;
    }
    lc=i;
    while((c[lc]==0)&&(lc>1)) lc--;
    for(i=lc;i>=1;i--) cout<<c[i];
    cout<<endl;
    return 0;
}

高精度乘法

//A*B problem.cpp
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char a1[2006],b1[2006];
int a[2006]={0},b[2006]={0},c[20000050]={0},lena,lenb,lenc,i,j,x;
int main(){
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    scanf("%s",a1);
    scanf("%s",b1);
    lena=strlen(a1);
    lenb=strlen(b1);
    for(i=0;i<lena;i++){
        a[lena-i]=a1[i]-'0';
    }
    for(i=0;i<lenb;i++){
        b[lenb-i]=b1[i]-'0';
    }
    for(i=1;i<=lena;i++){
        x=0;
        for(j=1;j<=lenb;j++){
            c[i+j-1]=a[i]*b[j]+x+c[i+j-1];
            x=c[i+j-1]/10;
            c[i+j-1]%=10;
        }
        c[i+lenb]=x;
    }
    lenc=lena+lenb;
    while(c[lenc]==0&&lenc>1){
        lenc--;
    }
    for(i=lenc;i>=1;i--){
        cout<<c[i];
    }
    cout<<endl;
    return 0;
}