hdu 1228java_hdu 1228 A + B -----模拟

A + B

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7852    Accepted Submission(s): 4417

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

3

90

96

直接 把字符串转化为数字。  遇到 '+'   换数字。  遇到  ‘=’  停止输入。

//Memory: 216 KBTime: 0 MS

//Language: C++Result: Accepted

#include

#include

int main()

{

char a[6][10];

int i,b[2],j,temp,ten,c;

while(scanf("%s",a[0])!=EOF)

{

ten=1;

c=0;

b[0]=b[1]=0;

for(i=0;;i++)

{

if(i!=0)

scanf("%s",a[i]);

if(a[i][0]=='=') break;

if(strcmp(a[i],"zero")==0)

temp=0;

if(strcmp(a[i],"one")==0)

temp=1;

if(strcmp(a[i],"two")==0)

temp=2;

if(strcmp(a[i],"three")==0)

temp=3;

if(strcmp(a[i],"four")==0)

temp=4;

if(strcmp(a[i],"five")==0)

temp=5;

if(strcmp(a[i],"six")==0)

temp=6;

if(strcmp(a[i],"seven")==0)

temp=7;

if(strcmp(a[i],"eight")==0)

temp=8;

if(strcmp(a[i],"nine")==0)

temp=9;

if(a[i][0]=='+')

{

ten=1;

c=1;

}

else

{

b[c]=b[c]*ten+temp;

ten*=10;

}

}

if(b[0]+b[1]==0) break;

printf("%d\n",b[0]+b[1]);

}

return 0;

}