Large Number Multiplication Algorithm

A few days ago, my senior colleague had an online coding test. Fearing she wouldn't have enough time, she asked me for help. I solved a few problems, and I felt they were commonly used in interviews. Large number multiplication was one of them, and I believe it will be frequently encountered in future interviews, so I made a note of it.

The method I adopted here is to store large numbers as strings, then multiply the two strings digit by digit, followed by carrying and shifting. There should be more efficient code available.

Source Code:

#include <stdio.h>#include <stdlib.h>#include <string.h> #define N 100 /* *将在数组中保存的字符串转成数字存到int数组中*/void getdigits(int *a,char *s){     int i;     char digit;     int len = strlen(s);      //对数组初始化     for(i = 0; i < N; ++i)           *(a + i) = 0;     for(i = 0; i < len; ++i){           digit = *(s + i);           *(a + len - 1 - i) = digit - '0';//字符串s="12345",因此第一个字符应该存在int数组的最后一个位置     }} /* *将数组a与数组b逐位相乘以后存入数组c */void multiply(int *a,int *b,int *c){     int i,j;      //数组初始化     for(i = 0; i < 2 * N; ++i)           *(c + i) = 0;	 /*	  *数组a中的每位逐位与数组b相乘，并把结果存入数组c	  *比如，12345*12345，a中的5与12345逐位相乘	  *对于数组c：*(c+i+j)在i=0,j=0,1,2,3,4时存的是5与各位相乘的结果	  *而在i=1,j=0,1,2,3,4时不光会存4与各位相乘的结果，还会累加上上次相乘的结果.这一点是需要注意的!!!	 */     for(i = 0; i < N; ++i)           for(j = 0; j < N; ++j)                 *(c + i + j) += *(a + i) * *(b + j);	 /*	  *这里是移位、进位	 */     for(i = 0; i < 2 * N - 1; ++i)     {           *(c + i + 1) += *(c + i)/10;//将十位上的数向前进位，并加上原来这个位上的数           *(c + i) = *(c + i)%10;//将剩余的数存原来的位置上     }} int main(){    int a[N],b[N],c[2*N];    char s1[N],s2[N];    int j = 2*N-1;    int i;        printf("input the first number:");    scanf("%s",s1);    printf("/ninput the second number:");    scanf("%s",s2);        getdigits(a,s1);    getdigits(b,s2);        multiply(a,b,c);        while(c[j] == 0)               j--;    for(i = j;i >= 0; --i)           printf("%d",c[i]);    printf("/n");    return 0;}