任意长度的两个正整数相乘

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两个大数相乘：char* multiply(char*,char*)。给了两个字符串，每个都是代表了一个很长的10进制表示的数， 比如 String str1 = "23456789009877666555544444"； String str2 = "346587436598437594375943875943875"， 最后求出它们的乘积。 不用jdk的数学运算包BigInteger.或者类似的包。

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MULTIPLY_SIZE 200

int m_atoi(char c);
char *multiply(char *s, char *muldstr, char *mulerstr);
char *reverse_string(char *str);

//ascii to char
int m_atoi(char c){
    return ((c>='0')&&(c<='9'))? (c-'0') : 0;
}

char *multiply(char *s, char *muldstr, char *mulerstr){
    int muldlen = strlen(muldstr); 
    int    mulerlen = strlen(mulerstr);

    int count = 0;
    int i, j;
    for(i=muldlen-1; i>=0; i--, count++){
        char *ptrs = s + count;  
        int carrier = 0;
        int muld = m_atoi(muldstr[i]);
        for(j=mulerlen-1; j>=0; j--){
            int mul, temp, mod;

            temp = (m_atoi(*ptrs)+carrier);

            mul = m_atoi(mulerstr[j]);
            temp = temp + muld * mul;
            
            *ptrs = temp % 10 + '0';
            carrier = temp/10;

            ptrs++;
        }
        if(carrier != 0){
            *ptrs = carrier + '0';
        }
    }
    reverse_string(s);
    return s;
}

char *reverse_string(char *s){
    int len = strlen(s);
    char *first = s;
    char *last = s + len -1;
                
    while(last > first){
        char c = *first;
        *first = *last;
        *last = c;
        
        first++;
        last--;
    }
    return s;
}

int main(int argc, char *argv[]){
    /*
    char s1[200];
    char s2[200];
    char *s;

    scanf_s("%s", s1, sizeof(s1));
    scanf_s("%s", s2, sizeof(s2));
    */
    
    char s1[200] = "23456789009877666555544444";
    char s2[200] = "346587436598437594375943875943875";
    char *s;
    
    s = (char *)malloc(sizeof(char)*MULTIPLY_SIZE);
    memset(s, 0, MULTIPLY_SIZE);
    multiply(s, s1, s2);

    printf("%s\n", s);
    free(s);

    system("pause");
    return 0;
}

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