高精度乘法

You'll be given two intergers. The number of the digits of each is from 1 to 200, inclusive.

    And what you need to do is to culcalate the product of the two intergers.

    The number of the digits of the product won't be more than 400.

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Input format:

    Two lines in total.

    The first line is an interger a(0 <= a < 10²⁰⁰), and the second line is an interger b(0 <= b < 10²⁰⁰).

Output format:

    Only one line, which is the product of a and b.

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Sample input:

12

2

Sample output:

24

 

 

 

1. #include<stdio.h>
2. #include<string.h>
3.  
4. int multiplication(char*a,char*b,int*c);
5. int main(){
6. int c[400+10], i;
7. int k;
8. char a[200+10]={0};
9. char b[200+10]={0};
10. memset(c,0,sizeof(c));
11. gets(a);
12. gets(b);
13. if((a[0]=='0')||(b[0]=='0')){
14. printf("0\n");
15. }else{
16. k = multiplication(a, b, c);
17. for(i = k; i >=1; i--) printf("%d", c[i]);
18. printf("\n");
19. }
20. return0;
21. }
22. int multiplication(char*a,char*b,int*c){
23. int i, j, m, tem, k =405, la, lb;
24. la = strlen(a);
25. lb = strlen(b);
26. for(i = lb -1; i >=0; i--){
27. for(j = la -1, m =0; j >=0; j--){
28. tem = c[(lb-i)+(la-j)-1];
29. c[(lb-i)+(la-j)-1]=
30. ((a[j]-'0')*(b[i]-'0')+ m + c[(lb - i)+(la - j)-1])%10;
31. m =((a[j]-'0')*(b[i]-'0')+ m + tem)/10;
32. }
33. c[(lb-i)+(la-j)-1]= m;
34. }
35. while(!c[k]) k--;
36. return k;
37. }

 

转载于:https://www.cnblogs.com/sysu-zhengwsh/p/3674201.html