2018 South University computer examination questions b

2018 South University computer examination questions b

Topic Link

Title Description

Reads two positive integer less than A and B 100 to calculate A + B, Note: A + each bit is given by the corresponding English letter B.

Entry

Test input comprising a plurality of test cases, each test case per line, the format of "A + B =", there is a space between two adjacent strings interval. When A and B are simultaneously input end is zero, do not output the corresponding results.

Export

An output line for each test case, i.e., the value of A + B.

Sample input

one + two =
three four + five six =
zero + zero =

Sample Output

3
90

Topic ideas

This question and talked before the word replace this idea is the same title, the word into a string replacement is about words, this question is the same. Then calculated. The Code

#include<stdio.h>
#include<string.h>
char digit[12][10] = {"zero","one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int judge(char *s)
{
    for(int i = 0; i < 10; i++)
    {
        if(!strcmp(digit[i], s))
            return i;
    }
}
int main()
{
    char str[100];
    int len, p;
    char word[20][20];
    int a, b, pow, temp, d[2][2], j;
    while(gets(str))
    {
        int l = strlen(str);
        p = len = 0;
        for(int i = 0; i < l; i++)
        {
            if(str[i] == '=')
            {
                continue;
            }
            if(str[i] == ' ')
            {
                word[len++][p] = '\0';
                p = 0;
            }
            else
                word[len][p++] = str[i];
        }
        pow = 1;
        p = 1;
        j = 0;
        memset(d, 0, sizeof(d));
        for(int i = len - 1; i >= 0; i--)
        {
            if(!strcmp(word[i], "+"))
            {
                p = 1;
                j++;
                continue;
            }
            d[j][p] = judge(word[i]);
            p--;
        }
        a = d[0][0] * 10 + d[0][1];
        b = d[1][0] * 10 + d[1][1];
        if(a == b && b == 0)
            break;
        printf("%d\n", a+b);
    }
    return 0;
}