两个大数（50位以上）相乘

废话也不多说，这个纯粹也是考一点点逻辑问题的，也很简单：

代码不是写的很好，也没有太多功夫去优化，请大牛勿喷。

#include<stdio.h>#define N 101//定义一个存放结果的字符数组char sum[2*N + 1];//将字符转换成数字int char_to_int(char ch){switch(ch){case '0':return 0;case '1':return 1;    case '2':return 2;case '3':return 3;case '4':return 4;case '5':return 5;case '6':return 6;case '7':return 7;case '8':return 8;case '9':return 9;}}//将数字转换成字符char int_to_char(int ch){switch(ch){case 0:return '0';case 1:return '1';case 2:return '2';case 3:return '3';case 4:return '4';case 5:return '5';case 6:return '6';case 7:return '7';case 8:return '8';case 9:return '9';}}//计算两个数之和void add(int k,int k2, char one[], char two[]){int i = k2;int max = k;int j = 0;int add_1[N + 1] = {0};  //用来判断是否有进位int c;//先计算两个有相同位数的部分   for(;i >= 0; max--,i--,j++) {c = char_to_int(one[max]) + char_to_int(two[i]) + add_1[j];if(c > 9) {sum[j] = int_to_char(c % 10);add_1[j + 1] = 1;}else {sum[j] = int_to_char(c);}}//再计算其中一个位数大于另外一个数的位数的部分i = k - k2 -1;for(;i >= 0; j++,i--,max--) {c = char_to_int(one[max]) + add_1[j];if(c > 9) {sum[j] = int_to_char(c % 10);add_1[j + 1] = 1;}else {sum[j] = int_to_char(c);}}//判断最后是否有进位if(add_1[j] == 1) {sum[j++] = '1';}    sum[j] = '\0';}//将字符数组反转,因为加法是从低位开始相加的void transvert(char str[],int n){char ch;int i;for(i = 0; i <= n / 2; i++) {ch = str[i];str[i] = str[n - i];str[n - i] = ch;}}//分别计算其中一个数的每一位与另外一个数的每一位相乘void add_to_twoarray(int k, int k2,char one[],char two[]){int i,j,c,m = 0;int line,over;char a[N + 1], b[2*N + 1]; //a[]放的是一位数和另外一个数每一位相乘的结果,                           //b[]放的是sum[]的结果int count_num;sum[0] = '0';sum[1] = '\0';for(i = k2; i >= 0; i--,m++) {int add_[N  + 1] = {0};for(line = 0; line < m; line++) {a[line] = '0';    //目的就是进行移位,因为每计算一次就会移位一次}for(j = k,over = 0; j >= 0; j--,line++,over++) {c = char_to_int(two[i]) * char_to_int(one[j]) + add_[over];add_[over + 1] = c / 10;a[line] = int_to_char(c % 10);}if(add_[over] > 0) {a[line++] = int_to_char(add_[over]);}a[line] = '\0';for(j = 0; sum[j] != '\0'; j++){b[j] = sum[j];}b[j] = '\0';count_num = j - 1;transvert(a,line-1);transvert(b,count_num);if(count_num > line -1) {add(count_num,line-1,b,a);}else {add(line-1,count_num,a,b);}}//最后求出结果的位数,然后进行反转输出for(j = 0; sum[j] != '\0'; j++){}transvert(sum,j -1);printf("%50s",sum);    printf("\n\n"); }//输入两个大数,判断哪个数比较大void two_bignumber_add(){char one[N],two[N];int i = 0,k,k2;printf("input a big number:\n");gets(one);printf("input another number:\n");gets(two);for(i = 0; one[i] != '\0'; i++){}k = i - 1;for(i = 0; two[i] != '\0'; i++){}k2 = i - 1;printf("\t\t--------------the result is:----------------");printf("\n%50s\n",one);printf("%50s\n",two);printf("\t\t--------------------------------------------\n");if(k >= k2) {add_to_twoarray(k,k2,one,two);}else if(k < k2) {add_to_twoarray(k2,k,two,one);}}int main(){two_bignumber_add();return 0;}