初学者不太看得懂程序

Question

 用英文单词模拟数学计算

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。
具体的输入输出格式规定如下：
输入格式：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zero时输入结束，相应的结果不要输出。
输出格式：对每个测试用例输出1行，即A+B的值。
输入样例：
one+ two =
three four +five six=
zero seven +eight nine =
zero + zero =
输出样例：
three
nine zero
nine six

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXSIZE (200)

char *array[] = {"zero", "one", "two", "three", "four", "five",
"six", "seven", "eight", "nine"};

void procstr(char *string, int *left, int *right)
{
char words[50][MAXSIZE];
char c;
int i, num = 0, word = 0, j;

/*
 *这部分的功能就是将字符串分解
 * 输入为 one + two =       
 * 分解为:        
 * words[0] = "one"
 * words[1] = "+"
 * words[2] = "two"
 * words[3] = "="
 */
for(i = 0; (c = string[i]) != '\0'; i++) {
    if (c == ' ') {
        if (word) {
            strncpy(words[num-1], &string[i-word], word);
            words[num-1][word] = '\0';
        }
        word = 0;
    } else if (word == 0) {
        word++;
        num++;
    } else {
        word++;
    }
}    

if (word) {
    strncpy(words[num-1], &string[i-word], word);
    words[num-1][word] = '\0';
}
/*****************************************/    

/*
 * 这里的功能就是把
 * + 左边的转换为整数放入 left 里面
 * + 右边的转换为整数放入 right 里面
 */

*left = 0 , *right = 0;
int leftok = 0;
for(i = 0; i<num - 1; i++) {
            if (strcmp(words[i], "+") == 0) {
        leftok = 1;
    } else {
        for (j = 0; j < 10; j++) {
            if (strcmp(words[i], array[j]) == 0) {
                if (leftok == 1) {
                    *right = (*right) * 10 + j;
                } else {
                    *left = (*left) * 10 + j;
                }
            }
        }
    }
}

}

/* 显示结果 */
void showRes(int res)
{
int stack[20], top = -1;

do {
    stack[++top] = res % 10;
    res /= 10;
} while (res != 0);

for (int i = top; i>=0; i--) {
    if (i != 0) {
        printf("%s ", array[stack[i]]);
    } else {
        printf("%s\n", array[stack[i]]);
    }
}

}

int main()
{
char str[MAXSIZE] = {'\0'};
int left = 0, right = 0;int sum;
while(1) {
fgets(str, MAXSIZE - 1, stdin);
procstr(str, &left, &right);

    if (left == 0 && right == 0) {
        break;
    } else {
    
        showRes(left + right);
    } 
}
system("pause");
return 0;

}

Answer 1

#include <stdio.h>
#include <string.h>
char chNumber[10][10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int fnStrToNum(char chNum[])
{
int i = 0;
for (i=0; i<10; i++)
{
if (0 == strcmp(chNum, chNumber[i]))
{
return i;
}
}
return 0;
}
int fnStrToInt(char chNum[][10], int n)
{
int i = 0;
int c = 0;
while (n--)
{
c *= 10;
c += fnStrToNum(chNum[i++]);
}
return c;
}
int fnIntToStr(char chNum[][10], int nNum)
{
int r = 0;
char chT[10];
if (nNum<10)
{
strcpy(chNum[0], chNumber[nNum]);
}
else
{
r = fnIntToStr(&chT, nNum/10);
strcpy(chNum[r], chT);
}
return r+1;
}
int main()
{
int i = 0;
int j = 0;
int k = 0;
char chA[2][10] = {0}, chB[2][10] = {0};
int iCount = 0; 
int iOut[1000]; 
while (1)
{
scanf("%s%s", chA[0], chA[1]);
if (chA[1][0] == '+')
{
scanf("%s%s", chB[0], chB[1]);
if (chB[1][0] == '=')
{
iOut[iCount] = fnStrToNum(chA[0]) + fnStrToNum(chB[0]);
}
else
{
iOut[iCount] = fnStrToNum(chA[0]) + fnStrToInt(chB, 2);
}
}
else
{
scanf("%s%s", chB[0], chB[1]);
if (chB[1][0] == '=')
{
iOut[iCount] = fnStrToInt(chA, 2) + fnStrToNum(chB[0]);
}
else
{
iOut[iCount] = fnStrToInt(chA, 2) + fnStrToInt(chB, 2);
}
}
if (!iOut[iCount])
{
break;
}
iCount++;
}
char chC[3][10];
for (i=0; i<iCount; i++)
{
k = fnIntToStr(chC, iOut[i]);
for (j=0; j<k; j++)
{
printf("%s ", chC[j]);
}
printf("\n");
}
return 0;
}

经过测试，感谢采纳！