【思考题】任意长度有理数乘法运算

目标需求：

//描述:
//长数相乘
//请编程实现：两个任意长度的数相乘，请输出结果.
//详细要求以及系统约束
//1）两个数可能是小数、整数、正数、负数；
//2）输入输出均为字符串形式，输入的字符串以“\0”结束，输出的结果字符串也必须以“\0”结束；
//3）输入的字符串不能是空字符串或非法字符串，否则返回-1，其他情况返回0；
//4）输出的结果字符串需要过滤掉整数位前以及小数位后无效的0，小数位为全0的，直接输出整数位；
//例1：相乘结果为11.345，此数值前后均不可以带0，“011.345”或者“0011.34500”等等前后带无效0的均视为错误输出。
//例2：080 × 0.125 结果是“10”
//5）输出的结果如果是正数或0需要过滤掉前面的+号。
//例1：相乘结果为+121，则“+121”为错误输出，“121”为正确输出。
//输入:
//输入两个任意长度的数
//输出:
//输出相乘的结果,如果输入的数不规范，输出结果为空
//样例输入:
//100000
//500000
//样例输出:
//50000000000

思路：

首先实现字符串加法，去除乘数以及被乘数的小数点，计算结果小数点位置，被乘数逐位与乘数相乘并进行相对的左移位(乘10)，然后对这些逐位相乘结果进行相加即可。

得到结果再进行截头截尾(去除结果字符串的开始的0串和小数末尾的0串)。

#include <stdio.h>#include <stdlib.h>#include <string.h>void str_z(char* s1, char* s2, char **s1z, char **s2z, int *sign){int s1sign=1,s2sign=1;if(s1[0]=='-'){s1sign=-1;*s1z = &s1[1];}else*s1z = s1;if(s2[0]=='-'){s2sign=-1;*s2z = &s2[1];}else*s2z = s2;*sign = s1sign*s2sign;}void zheng_xiao_fenli(char *s1, char* s2, int *xiaoshu_cnt, char **s1_num, char **s2_num){int s1len, s2len;int k1, k2, s1_xiaoshu_cnt=0, s2_xiaoshu_cnt=0;s1len = strlen(s1);s2len = strlen(s2);for(k1=s1len-1; k1>=0; k1--){if(s1[k1]>='0'&&s1[k1]<='9')continue;else if(s1[k1]=='.')s1_xiaoshu_cnt += s1len-k1-1;else{exit(0); //非法字符}}if(s1_xiaoshu_cnt==0){*s1_num = s1;}else{*s1_num = (char*)malloc(sizeof(char)*s1len);/*if(*s1_num==NULL){printf("malloc failed!\n");exit(0);}*/memcpy(*s1_num, s1, sizeof(char)*(s1len-s1_xiaoshu_cnt-1));memcpy(*s1_num+s1len-s1_xiaoshu_cnt-1, s1+s1len-s1_xiaoshu_cnt, sizeof(char)*s1_xiaoshu_cnt);*(*s1_num+s1len-1) = '\0';}for(k2=s2len-1; k2>=0; k2--){if(s2[k2]>='0'&&s2[k2]<='9')continue;else if(s2[k2]=='.')s2_xiaoshu_cnt += s2len-k2-1;else{exit(0); //非法字符}}if(s2_xiaoshu_cnt==0){*s2_num = s2;}else{*s2_num = (char*)malloc(sizeof(char)*s2len);/*if(*s2_num==NULL){printf("malloc failed!\n");exit(0);}*/memcpy(*s2_num, s2, sizeof(char)*(s2len-s2_xiaoshu_cnt-1));memcpy(*s2_num+s2len-s2_xiaoshu_cnt-1, s2+s2len-s2_xiaoshu_cnt, sizeof(char)*s2_xiaoshu_cnt);*(*s2_num+s2len-1) = '\0';}*xiaoshu_cnt = s1_xiaoshu_cnt+s2_xiaoshu_cnt;}char* str_add(char* s1, char* s2){int s1len,s2len,anslen;int k1,k2,ansk,c,s;char *ans,*ansrtn;s1len=strlen(s1);s2len=strlen(s2);anslen=s1len>s2len?s1len+1:s2len+1;ans = (char*)malloc(sizeof(char)*(anslen+1));/*if(ans==NULL){printf("malloc failed!\n");exit(0);}*/k1 = s1len-1;k2 = s2len-1;ansk = anslen-1;s=0; c=0;while(k1>=0||k2>=0){if(k1>=0&&k2>=0){c = s1[k1--]+s2[k2--]-'0'-'0'+s;ans[ansk--] = c%10+'0';s = c/10;}else if(k1>=0){c = s1[k1--]-'0'+s;ans[ansk--] = c%10+'0';s = c/10;}else if(k2>=0){c = s2[k2--]-'0'+s;ans[ansk--] = c%10+'0';s = c/10;}}ans[0] = s+'0';ans[anslen] = '\0';for(ansk=0; ansk<anslen; ansk++){if(ans[ansk]!='0')break;}if(ansk!=0){ansrtn = (char*)malloc(sizeof(char)*(anslen-ansk+1));if(ansrtn==NULL){printf("malloc failed!\n");exit(0);}strcpy(ansrtn, &ans[ansk]);free(ans);return ansrtn;}elsereturn ans;}char* str_mul(char* s1, char* s2){int s1len, s2len, k1, k2, k;char* mul_temp, *temp, *del_temp;int mul_temp_len;char* add_temp;s1len = strlen(s1);s2len = strlen(s2);add_temp = (char*)malloc(sizeof(char)*2);/*if(add_temp==NULL){printf("malloc failed!\n");exit(0);}*/add_temp[0] = '0';  add_temp[1] = '\0';for(k1=0; k1<s1len; k1++){mul_temp = (char*)malloc(sizeof(char)*2);/*if(mul_temp==NULL){printf("malloc failed!\n");exit(0);}*/mul_temp[0] = '0';  mul_temp[1] = '\0';if(s1[k1]=='0')continue;for(k=s1[k1]-'0'; k>0; k--){del_temp = mul_temp;mul_temp = str_add(mul_temp, s2);free(del_temp);}mul_temp_len = strlen(mul_temp);temp = (char*)malloc(sizeof(char)*(mul_temp_len+s1len-k1));/*if(temp==NULL){printf("malloc failed!\n");exit(0);}*/memset(temp, '0', sizeof(char)*(mul_temp_len+s1len-k1));memcpy(temp, mul_temp, sizeof(char)*mul_temp_len);temp[mul_temp_len+s1len-k1-1] = '\0';free(mul_temp);mul_temp = temp;add_temp = str_add(mul_temp, add_temp);}//printf("%s", add_temp);return add_temp;}int main(void){char s1[100];char s2[100];int sign, xs_cnt, k, kk, kp;char *s1z, *s2z;char *s1zheng, *s1xiao, *s2zheng, *s2xiao, *s1_num, *s2_num;char *ans; int anslen, xs_cnt_temp;gets(s1);gets(s2);str_z(s1, s2, &s1z, &s2z, &sign);zheng_xiao_fenli(s1z, s2z, &xs_cnt, &s1_num, &s2_num);ans = str_mul(s1_num, s2_num);anslen = strlen(ans);printf("\n");if(sign==-1)printf("-");if(xs_cnt==0){printf("%s", ans);}else{if(anslen<=xs_cnt){for(k=anslen-1; k>=0; k--){if(ans[k]!='0')break;}printf("0.");xs_cnt_temp = xs_cnt-anslen;while(xs_cnt_temp--){printf("0");}for(kk=0; kk<=k; kk++){printf("%c", ans[kk]);}}else{for(k=0; k<anslen-xs_cnt; k++){printf("%c", ans[k]);}for(kk=anslen-1; kk>=k; kk--){if(ans[kk]!='0')break;}if(kk!=k-1){printf(".");for(kp=k; kp<=kk; kp++){printf("%c", ans[kp]);}}}}system("pause");return 0;}

结果演示：