ZOJ 2971 Give Me the Number；ZOJ 2311 Inglish-Number Translator （字符处理，防空行，strstr）

ZOJ 2971 Give Me the Number 题目

ZOJ 2311 Inglish-Number Translator 题目

//两者题目差不多，细节有点点不一样，因为不是一起做的，所以处理方式有一点点不一样——还是前一个方式高端呀。

//ZOJ 2971 的AC代码（用了strstr函数）：

#include<stdio.h>
#include<string.h>
char c[][] = {"zero" , "one" , "two" , "three" , "four" , "five" , "six"  , "seven" , "eight" , "nine", "ten" ,
    "eleven" , "twelve" , "thirteen" , "fourteen" , "fifteen" , "sixteen" , "seventeen" , "eighteen" , "nineteen", "twenty" ,
    "thirty" , "forty" , "fifty" , "sixty" , "seventy" , "eighty" , "ninety" , "hundred" , "thousand" , "million" , "and"};
char s[],sss[];
int main()
{
    int ne , sum , flag , i,j , acc,len;
    int tt;
    scanf("%d",&tt);
    getchar();
    while(tt--)
    {
        gets(s);
        ne = ;
        sum = ;
        flag = ;
        acc = ;
        len=strlen(s);
        while (flag <len)
        {
            while (s[flag] == ' ') flag++;
            //    while (s[flag] == 'and') flag++;
            for (i =  ; i >=  ; i--)
            {
                if (strstr(&s[flag] , c[i]) == &s[flag])//在s中找c[i]如果找到了，返回位置，没有找到返回-1，即false；&取位置
                {
                    if (i <= ) acc += i;
                    else if (i <= ) acc += (i - ) * ;
                    else if (i == ) acc *= ;
                    else if (i == )
                    {
                        sum += acc * ;
                        acc = ;
                    }
                    else if (i == )
                    {
                        sum += acc * ;
                        acc = ;
                    }
                    else if (i == ) ne = -;
                    flag += strlen(c[i]);
                    break;
                }
            }
        }
        printf("%d\n" , (sum + acc));
 
    }
    return ;
}

字符串转化为数

//ZOJ 2311 的AC代码（这题要注意，防空行，就是一行什么都没输入，那么也什么都不要输出）：

//模拟
//英文句子转阿拉伯数字。
 
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
 
int main()
{
    int len,i,j,k,num,sign,ans;
    char str[],word[];
    char w[][]={"negative", "zero", "one", "two", "three", "four", "five",
        "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen",
        "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen",
        "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty",
        "ninety", "hundred", "thousand", "million"};
    while(gets(str))
    {
        if(strcmp(str,"")==)continue;//防空行
        len=strlen(str);
        str[len++]=' ';
        num=;
        j=;
        sign=;
        ans=;
        for(i=;i<len;i++)
        {
            if(str[i]!=' ')
                word[j++]=str[i];
            else
            {
                word[j++]='\0';
                for(k=;k<;k++)
                {
                    if(strcmp(w[k],word)==)
                    {
                        if(k==)
                            sign=-;
                        else if(k>=&&k<=)
                            num+=k-;
                        else if(k>=&&k<=)
                            num+=(k-)*;
                        else if(k==)
                            num=num*;
                        else if(k==)//只要考虑1000
                        {
                            ans+=num*;
                            num=;
                        }
                        else if(k==)//和1000000就可以了
                        {
                            ans+=num*;
                            num=;
                        }
                        break;
                    }
                }
                j=;
            }
        }
        ans+=num;
        printf("%d\n",ans*sign);
    }
    return ;
}