【模板】高精度计算
大数相加：
1 #include <stdio.h>
2 #include <string.h>
3 #include <stdlib.h>
4
5 void add(char a[],char b[],char back[])
6 {
7 int i,j,k,up,x,y,z,l;
8 char *c;
9 if(strlen(a) > strlen(b))
10 l = strlen(a)+2;
11 else
12 l = strlen(b)+2;
13 c = (char*)malloc(l*sizeof(char));
14 i = strlen(a)-1;
15 j = strlen(b)-1;
16 k = 0;
17 up = 0;
18 while(j>=0 || i>=0)
19 {
20 if(i<0) x = ‘0‘;
21 else
22 x = a[i];
23 if(j<0) y = ‘0‘;
24 else
25 y = b[j];
26 z = x-‘0‘+y-‘0‘;
27 if(up)
28 z++;
29 if(z>9)
30 {
31 up = 1;
32 z%=10;
33 }
34 else
35 up = 0;
36 c[k++] = z+‘0‘;
37 i--;
38 j--;
39 }
40 if(up)
41 c[k++] = ‘1‘;
42 i = 0;
43 c[k] = ‘\0‘;
44 for(k-=1; k>=0; k--)
45 back[i++] = c[k];
46 back[i] = ‘\0‘;
47 }
48
49 int main()
50 {
51 char c[10000],t[10000],sum[1000];
52 int m;
53 scanf("%d%*c",&m);
54 while(m--)
55 {
56 scanf("%s%s",c,t);
57 add(c,t,sum);
58 printf("%s\n",sum);
59 if(m)
60 printf("\n");
61 }
62 return 0;
63 }
大数加小数：
1 void addt(char a[],int b,char c[])
2 {
3 int len=strlen(a);
4 char s[100];
5 for(int i=0; i<len; i++)
6 s[len-i-1]=a[i]-‘0‘;
7 int add=0;
8 for(int i=0;; i++)
9 {
10 if(i>=len)
11 {
12 s[i]=0;
13 len++;
14 }
15 int k=s[i]+b%10+add;
16 b/=10;
17 if(k>=10)
18 {
19 s[i]=k%10;
20 add=k/10;
21 }
22 else
23 {
24 s[i]=k;
25 add=0;
26 }
27 if(b==0&&add==0)break;
28 }
29 for(int i=0; i<len; i++)
30 c[len-1-i]=s[i]+‘0‘;
31 c[len]=‘\0‘;
32
33 }
34
35 int main()
36 {
37 char c[1000],t[1000],sum[1000];
38 int m;
39 while(~scanf("%s%d",c,&m))
40 {
41 mult(c,m,t);
42 printf("%s\n",t);
43 addt(c,m,sum);
44 printf("%s\n",sum);
45 }
46 return 0;
47 }
大数相乘：
1 #include <stdio.h>
2 #include <string.h>
3
4 void mult(char a[],char b[],char s[])
5 {
6 int i,j,k = 0,alen,blen,sum = 0,res[65][65]={0},flag = 0;
7 char result[65];
8 alen = strlen(a);
9 blen = strlen(b);
10 for(i = 0;i<alen;i++)
11 {
12 for(j = 0;j<blen;j++)
13 res[i][j] = (a[i]-‘0‘)*(b[j]-‘0‘);
14 }
15 for(i = alen-1;i>=0;i--)
16 {
17 for(j = blen-1;j>=0;j--)
18 {
19 sum = sum+res[i+blen-j-1][j];
20 printf("res = %d\n",res[i+blen-j-1][j]);
21 }
22 result[k] = sum%10;
23 k++;
24 sum = sum/10;
25 }
26 for(i = blen-2;i>=0;i--)
27 {
28 for(j = 0;j<=i;j++)
29 {
30 sum = sum+res[i-j][j];
31 }
32 result[k] = sum%10;
33 k++;
34 sum = sum/10;
35 }
36 if(sum)
37 {
38 result[k] = sum;
39 k++;
40 }
41 for(i = 0;i<k;i++)
42 result[i]+=‘0‘;
43 for(i = k-1;i>=0;i--)
44 s[i] = result[k-1-i];
45 s[k] = ‘\0‘;
46 while(1)
47 {
48 if(strlen(s)!=strlen(a) && s[0] == ‘0‘)
49 strcpy(s,s+1);
50 else
51 break;
52 }
53 }
54
55 int main()
56 {
57 char c[1000],t[1000],sum[1000];
58 int m;
59 while(~scanf("%s%s",c,t))
60 {
61 mult(c,t,sum);
62 printf("%s\n",sum);
63 }
64 return 0;
65 }
大数乘小数：
1 #include <stdio.h>
2 #include <string.h>
3
4 void mult(char c[],int m,char t[])
5 {
6 char s[100];
7 int len=strlen(c);
8 for(int i=0; i<len; i++)
9 s[len-i-1]=c[i]-‘0‘;
10 int flag,add=0;
11 for(int i=0; i<len; i++)
12 {
13 int k=s[i]*m+add;
14 if(k>=10)
15 {
16 s[i]=k%10;
17 add=k/10;
18 flag=1;
19 }
20 else
21 {
22 s[i]=k;
23 add=0;
24 flag=0;
25 }
26 }
27 while(add)
28 {
29 s[len++]=add%10;
30 add/=10;
31 }
32 for(int i=0; i<len; i++)
33 t[len-1-i]=s[i]+‘0‘;
34 t[len]=‘\0‘;
35 }
大数相乘：
1 #include <stdio.h>
2 #include <string.h>
3
4 void sub(char s1[],char s2[],char t[])
5 {
6 int i,l1,l2,k;
7 l2 = strlen(s2);
8 l1 = strlen(s1);
9 t[l1] = ‘\0‘;
10 l1--;
11 for(i = l2-1; i>=0; i--,l1--)
12 {
13 if(s1[l1]-s2[i]>=0)
14 t[l1] = s1[l1] - s2[i] + ‘0‘;
15 else
16 {
17 t[l1] = 10+s1[l1] - s2[i]+‘0‘;
18 s1[l1-1]= s1[l1-1] -1;
19 }
20 }
21 k = l1;
22 while(s1[k]<0)
23 {
24 s1[k]+=10;
25 s1[k-1]-=1;
26 k--;
27 }
28 while(l1>=0)
29 {
30 t[l1] = s1[l1];
31 l1--;
32 }
33 loop:
34 if(t[0] == ‘0‘)
35 {
36 l1 = strlen(s1);
37 for(i = 0; i<l1-1; i++)
38 t[i] = t[i+1];
39 t[l1-1] = ‘\0‘;
40 goto loop;
41 }
42 if(strlen(t) == 0)
43 {
44 t[0] = ‘0‘;
45 t[1] = ‘\0‘;
46 }
47 }
48
49 int main()
50 {
51 char c[1000],t[1000],sum[1000];
52 int m;
53 while(~scanf("%s%s",c,t))
54 {
55 sub(c,t,sum);
56 printf("%s\n",sum);
57 }
58 return 0;
59 }
大数阶乘：
1 #include<iostream>
2 #include<cmath>
3 using namespace std;
4 int main()
5 {
6 void factorial(int n,int *) ;
7 int b[10000];
8 int n;
9 while(cin>>n)
10 {
11 factorial(n,b);
12 }
13 return 0;
14 }
15 void factorial(int n,int b[10000])
16 {
17 long a[10000];
18 int i,j,l,c,m=0,w;
19 a[0]=1;
20 for(i=1; i<=n; i++)
21 {
22 c=0;
23 for(j=0; j<=m; j++)
24 {
25 a[j]=a[j]*i+c;
26 c=a[j]/10000;
27 a[j]=a[j]%10000;
28 }
29 if(c>0)
30 {
31 m++;
32 a[m]=c;
33 }
34 }
35 w = m*4+log10(a[m])+1;
36 cout << w << endl;
37 cout<<a[m];
38 for(i=m-1; i>=0; i--)
39 cout<<a[i];
40 cout<<endl;
41 }
大数进制转换：
1 #include <stdio.h>
2 #include <string.h>
3
4 void consversion(char s[],char s2[],long d1,long d2)
5 {
6 long i,j,t,num;
7 char c;
8 num = 0;
9 for(i = 0;s[i]!=‘\0‘;i++)
10 {
11 if(s[i]<=‘9‘ && s[i]>=‘0‘)
12 t = s[i] - ‘0‘;
13 else
14 t = s[i] - ‘A‘ +10;
15 num = num*d1+t;
16 }
17 i = 0;
18 while(1)
19 {
20 t = num%d2;
21 if(t<=9)
22 s2[i] = t+‘0‘;
23 else
24 s2[i] = t+‘A‘-10;
25 num/=d2;
26 if(num == 0)
27 break;
28 i++;
29 }
30 for(j = 0;j<i/2;j++)
31 {
32 c = s2[j];
33 s2[j] = s[i-j];
34 s2[i-j] = c;
35 }
36 s2[i+1]=‘\0‘;
37 }
38
39 int main()
40 {
41 char s1[1000],s2[1000];
42 int d1,d2;
43 while(~scanf("%s%d%d",s1,&d1,&d2))
44 {
45 consversion(s1,s2,d1,d2);
46 //将d1进制的s1转换成d2进制s2.
47 printf("%s\n",s2);
48 }
49
50 return 0;
51 }
时间： 2024-11-05 17:29:25