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  • CDZSC_2015寒假新人(1)——基础 G

    Description

    Ignatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them.       
                  

    Input

    The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.        Each test case contains a single line with several words. There will be at most 1000 characters in a line.       
                  

    Output

    For each test case, you should output the text which is processed.       
                  

    Sample Input

    3 olleh !dlrow
    m'I morf .udh
    I ekil .mca
                  

    Sample Output

    hello world!
    I'm from hdu.
    I like acm.

    Hint

    Remember to use getchar() to read ' ' after the interger T, then you may use gets() to read a line and process it.

     

    思路:开两个字符数组,一个用来输入,一个用来存每一个单词(遇到空格停止倒着输出),注意最后一个单独输出

     

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    int main()
    {
    #ifdef CDZSC_OFFLINE
        freopen("in.txt","r",stdin);
    #endif
        char str[1024],s[1024];
        int len,i,j,t,p,flag;
        scanf("%d",&t);
        getchar();
        while(t--)
        {
            cin.getline(str,1024);
            len=strlen(str);
            p=0,flag=0;
            for(i=0; i<len; i++)
            {
                if(str[i]!=' ')
                {
                    s[p++]=str[i];
                }
                else
                {
                    if(flag==1)
                    {
                        printf(" ");
                    }
                    for(j=p-1; j>=0; j--)
                    {
                        printf("%c",s[j]);
                    }
                    flag=1;
                    p=0;
                }
            }
            printf(" ");  
            for(i=p-1; i>=0; i--)
            {
                printf("%c",s[i]);  
            }
            printf("
    ");
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/Wing0624/p/4243833.html
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