1412201333-hd-A + B

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A + B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12596    Accepted Submission(s): 7378


Problem Description
读入两个小于100的正整数A和B,计算A+B.
需要注意的是:A和B的每一位数字由对应的英文单词给出.
 

Input
测试输入包含若干测试用例,每个测试用例占一行,格式为"A + B =",相邻两字符串有一个空格间隔.当A和B同时为0时输入结束,相应的结果不要输出.
 

Output
对每个测试用例输出1行,即A+B的值.
 

Sample Input
one + two =three four + five six =zero seven + eight nine =zero + zero =
 

Sample Output
39096
 解题思路
        用gets()!=NULL输入并存储,然后判断a到‘+’截至,然后判断b到‘=’截至,最后输出a+b之和。
代码
#include<stdio.h>#include<string.h>#include<string.h>char s[1100];char num[1100];int switcha(char c[]){if(strcmp(c,"zero")==0)     return 0;else if(strcmp(c,"one")==0)    return 1;else if(strcmp(c,"two")==0)    return 2;else if(strcmp(c,"three")==0)    return 3;else if(strcmp(c,"four")==0)    return 4;else if(strcmp(c,"five")==0)    return 5;else if(strcmp(c,"six")==0)    return 6;else if(strcmp(c,"seven")==0)    return 7;else if(strcmp(c,"eight")==0)    return 8;else if(strcmp(c,"nine")==0)    return 9; }int main(){int a,b;int i,j,k;int len;while(gets(s)){len=strlen(s);a=0;for(i=0,j=0;s[i]!='+';i++){if(s[i]!=' '){num[j]=s[i];j++;}else{a=a*10+switcha(num);j=0;memset(num,0,sizeof(num));//注意用完之后清零,避免这一次three,写一次one结果存储为oneee的情况 }}k=i+2;b=0;for(i=k,j=0;s[i]!='=';i++){if(s[i]!=' '){num[j]=s[i];j++;}else{b=b*10+switcha(num);j=0;memset(num,0,sizeof(num));}}if(a==0&&b==0)    break;printf("%d\n",a+b);}return 0;} 


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