两位字符串加法

字符串加法程序解析

最新推荐文章于 2019-12-21 19:49:45 发布

转载最新推荐文章于 2019-12-21 19:49:45 发布 · 117 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/huashiyiqike/articles/2861913.html

#include <string.h> 
#include <stdio.h> 
#include <stdlib.h>
#include<iostream>
#include<math.h>
using namespace std;
char num[10][10]={"zero","one","two","three","four","five","six","seven","eight"
,"nine"};
int main(void) 
 { 
    char a[1000],b[1000],c[1000],e[5],f[5];int len,middle,temp,i,j,sumb,sumc,sum[3],sumall,temp2;
    while(cin.getline(a,1000,'\n'))
    {
        char add='+';
        char empty=' ';
        int len=strlen(a);
        len-=2;//a[len+1]='\0';
        int middle=(strchr(a,add)-a);
        for(i=0;i<middle;i++)
        {
            b[i]=a[i];
        }
        b[i]='\0';
        i+=2;j=0;
        for(;i<len+1;i++)
        {
            c[j++]=a[i];
        }
        c[j]='\0';
        temp=0;
        for(i=0;i<4;i++)
                e[i]=b[i];
            e[i]='\0';
        for(i=0;i<4;i++)
                f[i]=b[i];
            f[i]='\0';
        if(strcmp(e,num[0])==0&&strcmp(f,num[0])==0)
            return 0;
        sumall=0;
        memset(sum,0,sizeof(int)*3);
        while(strchr(b,empty))
        {
            int middle=(strchr(b,empty)-b);
            for(i=0;i<middle;i++)
                a[i]=b[i];
            a[i]='\0';
            for(i=0;i<10;i++)
                if(strcmp(a,num[i])==0)
                    sum[temp++]=i;
                temp2=strlen(b);
            for(i=0;i<=temp2-middle;i++)
                b[i]=b[i+middle+1];

        }
        temp--;
        for(i=temp;i>=0;i--)
            sumall+=sum[i]*pow(10,(temp-i));
        memset(sum,0,sizeof(int)*3);
        temp=0;
        while(strchr(c,empty))
        {
            int middle=(strchr(c,empty)-c);
            for(i=0;i<middle;i++)
                a[i]=c[i];
            a[i]='\0';
            for(i=0;i<10;i++)
                if(strcmp(a,num[i])==0)
                    sum[temp++]=i;
                temp2=strlen(c);
          for(i=0;i<=temp2-middle;i++) c[i]=c[i+middle+1];
        }
        temp--;
        for(i=temp;i>=0;i--)
            sumall+=sum[i]*pow(10,(temp-i));
        printf("%d\n",sumall);
        

        
    }
 }

转载于:https://www.cnblogs.com/huashiyiqike/articles/2861913.html