大数运算

最近在看《挑战编程》的高精度，又学到了很多。但是，单从实用性来说，它的实现方式复杂了，而这里就要求更容易在比赛中实现的模版。

大数加法，请参照本博客的一篇题解。

大数减法：

void subtract_big(char *a, char *b, char *c){
    int i, j, flag=0;
    int len1 = strlen(a), len2 = strlen(b);
    char *pmax = a, *pmin = b;

    memset(c, 0, sizeof(c)*max(len1, len2)+3);

    if(len2 > len1 || (len1 == len2 && strcmp(a, b) < 0)){
        pmax = b; pmin = a;
        flag = 1;
    }
    len1 = strlen(pmax); len2 = strlen(pmin);

    for(i=len1-1, j=len2-1; i>=0; i--, j--){
        if(j>=0) c[i+1] = pmax[i] - pmin[j] + '0';
        else c[i+1] = pmax[i];
    }
    if(flag) c[0] = '-';

    for(i=len1; i>=1; i--) if(c[i] < '0'){c[i] += 10; c[i-1]--;}

    i = 1;
    while(c[i] == '0') i++;
    if(c[i] == '\0'){
        c[0] = '0'; c[1] = '\0';
    }
    else{
        if(flag) j = 1;
        else j = 0;

        while(i<=len1){
            c[j++] = c[i++];
        }
        c[j] = '\0';
    }
}

大数乘除（这里除法并非大数除以大数）：

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

void multiply_big(char *a, char *b, char *c){//大数乘法
    int len1 = strlen(a), len2 = strlen(b), *s;
    int i, j;

    s = (int *)malloc(sizeof(int)*len1*len2);

    for(i=0; i<len1+len2; i++) s[i] = 0;

    for(i=0; i<len1; i++){
        for(j=0; j<len2; j++){
            s[i+j+1] += (a[i]-'0')*(b[j]-'0');
        }
    }

    for(i=len1+len2-1; i>=1; i--){
        if(s[i] >= 10){
            s[i-1] += s[i] / 10;
            s[i] %= 10;
        }
    }

    i=0;
    while(s[i] == 0) i++;
    for(j=0; i<len1+len2; i++, j++) c[j] = s[i] + '0';

    c[j] = '\0';

    free(s);
}

int dividor_big(char *a, int b, char *c){   //大数除法，这里除数并非大数，返回余数
    char *s;
    int tmp, i, j, len = strlen(a);

    s = (char *)malloc(sizeof(char)*(len+1));

    tmp = 0;
    for(i=0; i<len; i++){
        tmp = tmp*10+a[i]-'0';
        s[i] = tmp / b + '0';
        tmp %= b;
    }

    s[i] = '\0';
    for(i=0; s[i] == '0' && s[i] != '\0'; i++);

    if(s[i] == '\0'){
        c[0] = '0';
        c[1] = '\0';
    }
    else{
        for(j=0; i<len; i++, j++) c[j] = s[i];
        c[j] = '\0';
    }

    free(s);
    return tmp;
}

int main(){
    freopen("d:\\my.txt", "r", stdin);
    //char a[1000], b[1000], c[1000];
    char a[1000], c[1000];
    int b;

    //scanf("%s %s", a, b);
    scanf("%s %d", a, &b);
   // multiply_big(a, b, c);
    dividor_big(a, b, c);
    printf("%s\n", c);

    return 0;
}