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【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 }
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原文地址：http://www.cnblogs.com/jeff-wgc/p/4473388.html
