九度oj 1010 A+B

题目描述：

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.

输入：

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.

输出：

对每个测试用例输出1行,即A+B的值.

样例输入：

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

样例输出：

3
90
96
这本来是一到水题，但一开始不太会用strcmp函数，又知道scanf(%s)会按空格分开读，所以一开始写出了如下的代码

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>

#define HIGH 11
#define LEN 8
#define MAX 102
using namespace std;
char s[HIGH][LEN] = {"zero","one","two","three","four","five","six","seven","eight","nine"};

char source[MAX];
char A[MAX];
char B[MAX];
int main(int argc, char const *argv[])
{
    int a,b;
    while(gets(source)) {
        int i = 0;
        a = 0;
        b = 0;
        int state = 0;
        while(i < strlen(source)) {
            if(source[i] == '+') {
                state = 1;
                i++;
                continue;
            }
            if(source[i] == ' ' || source[i] == '=') {
                i++;
                continue;
            }
            for(int j = 0; j < 10; j++) {
                if(source[i] == s[j][0]) {
                    if(source[i] == 't') {
                        if(source[i+1] == 'w') {
                            j = 2;
                        }
                        else if(source[i+1] == 'h') {
                            j = 3;
                        }
                    }
                    else if(source[i] == 'f') {
                        if(source[i+1] == 'o') {
                            j = 4;
                        }
                        else if(source[i+1] == 'i') {
                            j = 5;
                        }
                    }
                    else if(source[i] == 's') {
                        if(source[i+1] == 'e') {
                            j = 7;
                        }
                        else if(source[i+1] == 'i') {
                            j = 6;
                        }
                    }
                    if(state == 0) {
                        a = 10 * a + j;
                    }
                    else if(state == 1) {
                        b = 10 * b + j;
                        
                    }
                    i = i + strlen(s[j]);
                    break;
                }
                
            }
            
        }
        if(a == 0 && b == 0) {
            break;
        }
        int c = a + b;
        printf("%d
",c);
    }
    return 0;
}

但后来发现，scanf的这个性质其实可以利用起来，所以写出如下的代码

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>

#define HIGH 11
#define LEN 8
#define MAX 102
using namespace std;
char s[HIGH][LEN] = {"zero","one","two","three","four","five","six","seven","eight","nine"};

int main(int argc, char const *argv[])
{
    char temp[MAX];
    int state = 0;
    int a = 0, b = 0;
    while(scanf("%s",temp)) {
        if(strcmp(temp,"+") == 0) {
            state = 1;
            continue;
        }

        if(strcmp(temp,"=") == 0) {
            if(a == 0 && b == 0) {
                break;
            }
            else {
                printf("%d
",a+b);
                state = 0;
                a = 0;
                b = 0;
                continue;
            }
        }
        for(int i = 0; i < 10; i++) {
            if(strcmp(temp,s[i])== 0) {
                if(state == 0) {
                    a = a * 10 + i;
                }
                else {
                    b = b * 10 + i;
                }
                continue;
            }
        }

    }

    return 0;
}

需要注意的是，strcmp这个函数，

若str1=str2，则返回零；

若str1<str2，则返回负数；

若str1>str2，则返回正数。

所以判断是否相等要判断其值是否为0