[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

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 char a1[500001],b1[500001];
 6 int a[500001],b[500001],c[500001],x,lenc,la,lb,i;
 7 int main(){
 8     memset(a,0,sizeof(a));
 9     memset(b,0,sizeof(b));
10     memset(c,0,sizeof(c));
11     gets(a1);
12     gets(b1);
13     la=strlen(a1);
14     lb=strlen(b1);
15     for(i=0;i<=la-1;i++){
16         a[la-i]=a1[i]-'0';
17     }
18     for(i=0;i<=lb-1;i++){
19         b[lb-i]=b1[i]-'0';
20     }
21     lenc=1;
22     x=0;
23     while(lenc<=la||lenc<=lb){
24         c[lenc]=a[lenc]+b[lenc]+x;
25         x=c[lenc]/10;
26         c[lenc]%=10;
27         lenc++;
28     }
29     c[lenc]=x;
30     if(c[lenc]==0) lenc--;
31     for(i=lenc;i>=1;i--) cout<<int(c[i]);
32     return 0;
33 }
A+B Problem

高精度减法

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 int main(){
 6     int a[10001],b[10001],c[10001],la,lb,lc,i;
 7     char n[10001],n1[10001],n2[10001];
 8     memset(a,0,sizeof(a));
 9     memset(b,0,sizeof(b));
10     memset(c,0,sizeof(c));
11     cin>>n1>>n2;
12     if(strlen(n1)<strlen(n2)||(strlen(n1)==strlen(n2)&&strcmp(n1,n2)<0)){
13         strcpy(n,n1);
14         strcpy(n1,n2);
15         strcpy(n2,n);
16         cout<<"-";
17     }
18     la=strlen(n1);
19     lb=strlen(n2);
20     for(i=0;i<=la-1;i++){
21         a[la-i]=int(n1[i]-'0');
22     }
23     for(i=0;i<=lb-1;i++){
24         b[lb-i]=int(n2[i]-'0');
25     }
26     i=1;
27     while(i<=la||i<=lb){
28         if(a[i]<b[i]){
29             a[i]+=10;
30             a[i+1]--;
31         }
32         c[i]=a[i]-b[i];
33         i++;
34     }
35     lc=i;
36     while((c[lc]==0)&&(lc>1)) lc--;
37     for(i=lc;i>=1;i--) cout<<c[i];
38     cout<<endl;
39     return 0;
40 }
A-B Problem

高精度乘法

 1 //A*B problem.cpp
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 using namespace std;
 6 char a1[2006],b1[2006];
 7 int a[2006]={0},b[2006]={0},c[20000050]={0},lena,lenb,lenc,i,j,x;
 8 int main(){
 9     memset(a,0,sizeof(a));
10     memset(b,0,sizeof(b));
11     memset(c,0,sizeof(c));
12     scanf("%s",a1);
13     scanf("%s",b1);
14     lena=strlen(a1);
15     lenb=strlen(b1);
16     for(i=0;i<lena;i++){
17         a[lena-i]=a1[i]-'0';
18     }
19     for(i=0;i<lenb;i++){
20         b[lenb-i]=b1[i]-'0';
21     }
22     for(i=1;i<=lena;i++){
23         x=0;
24         for(j=1;j<=lenb;j++){
25             c[i+j-1]=a[i]*b[j]+x+c[i+j-1];
26             x=c[i+j-1]/10;
27             c[i+j-1]%=10;
28         }
29         c[i+lenb]=x;
30     }
31     lenc=lena+lenb;
32     while(c[lenc]==0&&lenc>1){
33         lenc--;
34     }
35     for(i=lenc;i>=1;i--){
36         cout<<c[i];
37     }
38     cout<<endl;
39     return 0;
40 }
A*B Problem

 

posted @ 2018-07-04 08:53  Steve_braveman  阅读(108)  评论(0)  编辑  收藏  举报