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  • One-Two-Three

    Your little brother has just learnt to write one, two and three, in English. He has written a lot of those words in a paper, your task is to recognize them. Note that your little brother is only a child, so he may make small mistakes: for each word, there might be at most one wrong letter. The word length is always correct. It is guaranteed that each letter he wrote is in lower-case, and each word he wrote has a unique interpretation.

     

    Input 

    The first line contains the number of words that your little brother has written. Each of the following lines contains a single word with all letters in lower-case. The words satisfy the constraints above: at most one letter might be wrong, but the word length is always correct. There will be at most 10 words in the input.

     

    Output 

    For each test case, print the numerical value of the word.

     

    Sample Input 

     

    3
    owe
    too
    theee
    

     

    Sample Output 

     

    1
    2
    3

    分析:
    相当于字符和数字间的转换,已知长度没有错误,每个字符最多错一个地方

    代码:
    #include<stdio.h>
    #include<string.h>
    int main()
    {
        int n,l,i;
        char s[10];
        scanf("%d",&n);
            while(n--)
            {
                scanf("%s",s);
                l=strlen(s);
                if(l==5)
                   printf("3
    ");
                else
                {
                    if(s[0]=='o'&&s[1]=='n'||s[0]=='o'&&s[2]=='e'||s[1]=='n'&&s[2]=='e')
                       printf("1
    ");
                    else
                       printf("2
    ");
                }
            }
        
        return 0;
    }

    参考代码:

    #include<set>
    #include<map>
    #include<stack>
    #include<queue>
    #include<ctime>
    #include<cmath>
    #include<vector>
    #include<string>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #define Master Love
    #define YYF Handsome
    #define Rp Maxunsignedlonglongint
    #define Orz zhanhb8 qs1994
    #define Designer Zero
    #define zhx zh[x]
    #define lth th<<1
    #define rth th<<1|1
    #define middle (nl+nr)>>1
    #define MAX(a,b)  (a>b)?a:b
    #define MIN(a,b)  (a<b)?a:b
    #define getlen(s)  strlen(s+1)
    #define outputans cout<<ans<<endl
    #define inputint(k) scanf("%d",&k)
    #define inputchar(c) scanf("%c",&c)
    #define input64(k) scanf("%I64d",&k)
    #define inputdouble(k) scanf("%lf",&k)
    #define inputstring(s) scanf("%s",s+1)
    #define forout(i,l,r) for(i=l;i<=r;i++)
    #define forin(i,l,r) for(int i=l;i<=r;i++)
    #define clearit(a,k) memset(a,k,sizeof(a))
    #define formatrix for(int i=1;i<=m;i++)for(int j=1;j<=n;j++)
    #pragma warning(disable:4996)
    using namespace std;
    int T;
    char s[521];
    int main(){
        cin >> T;
        while (T--){
            inputstring(s);
            int len = strlen(s + 1);
            if (len == 5)printf("3
    ");
            else {
                int cnt = 0;
                if (s[1] == 't')cnt++;
                if (s[2] == 'w')cnt++;
                if (s[3] == 'o')cnt++;
                if (cnt >= 2)printf("2
    ");
                else printf("1
    ");
            
            }
        
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lipching/p/3854609.html
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