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

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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 ;

}