首页 > ACM题库 > HDU-杭电 > HDU 2816-I Love You Too-模拟-[解题报告]HOJ
2014
02-17

HDU 2816-I Love You Too-模拟-[解题报告]HOJ

I Love You Too

问题描述 :

This is a true story. A man showed his love to a girl,but the girl didn’t replied clearly ,just gave him a Morse Code:
****-/*—-/—-*/****-/****-/*—-/—**/*—-/****-/*—-/-****/***–/****-/*—-/—-*/**—/-****/**—/**—/***–/–***/****-/   He was so anxious that he asked for help in the Internet and after one day a girl named "Pianyi angel" found the secret of this code. She translate this code as this five steps:
1.First translate the morse code to a number string:4194418141634192622374
2.Second she cut two number as one group 41 94 41 81 41 63 41 92 62 23 74,according to standard Mobile phone can get this alphabet:GZGTGOGXNCS

3.Third she change this alphabet according to the keyboard:QWERTYUIOPASDFGHJKLZXCVBNM = ABCDEFGHIJKLMNOPQRSTUVWXYZ
So ,we can get OTOEOIOUYVL
4.Fourth, divide this alphabet to two parts: OTOEOI and OUYVL, compose again.we will get OOTUOYEVOLI
5.Finally,reverse this alphabet the answer will appear : I LOVE YOU TOO

I guess you might worship Pianyi angel as me,so let’s Orz her.
Now,the task is translate the number strings.

输入:

A number string each line(length <= 1000). I ensure all input are legal.

输出:

A number string each line(length <= 1000). I ensure all input are legal.

样例输入:

4194418141634192622374
41944181416341926223

样例输出:

ILOVEYOUTOO
VOYEUOOTIO

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2816

题目大意:

给你一串数字,通过各种转换得到一个序列。。。。。

解题思路:

水题一道,但是自己代码能力太弱,处理字符串常常力不从心。。看来以后得全面开战,不能老刷一个专题了。。。改变一下。

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
char num[9][4] = {{' ', ' ', ' ', ' '}, {'A', 'B', 'C', ' '},{'D', 'E', 'F', ' '},{'G', 'H', 'I', ' '},{'J', 'K', 'L', ' '},{'M', 'N', 'O',' '},{'P', 'Q', ' R', 'S'},{'T', 'U', 'V', ' '},{'W', 'X', 'Y', 'Z'}};
string a = "QWERTYUIOPASDFGHJKLZXCVBNM";
string b = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

char fun(char c)
{
    for(int i = 0; i < 52; ++i)
        if(a[i] == c)
            return b[i];
}

int main()
{
    char p[1010];
    string s, t, qian, hou, res;
    bool flag;
    int len, t1;
    while(~scanf("%s", p))
    {
        len = strlen(p);
        flag = true;
        s.clear(); t.clear(); qian.clear(); hou.clear(); res.clear();
        for(int i = 0; i < len; i += 2)
            s += num[(p[i] - '0') - 1][(p[i + 1] - '0') - 1];
        len = s.length();
        t1 = len;
        for(int i = 0; i < len; ++i)
            t += fun(s[i]);
        qian = t.substr(0, (len + 1) / 2);
        hou = t.substr((len + 1) / 2);
        len = qian.length();
        int len1 = hou.length();
        for(int i = 0, j = 0; i < len && j < len1; ++i, ++j)
        {
            res += qian[i];
            res += hou[j];
        }
        if(t1 % 2)
            res += qian[len - 1];
        for(int i = len + len1 - 1; i >= 0; --i)
            cout<<res[i];
        cout<<endl;
    }
    return 0;
}

解题参考:http://blog.csdn.net/niushuai666/article/details/7184441


  1. #include <cstdio>
    #include <cstring>

    const int MAXSIZE=256;
    //char store[MAXSIZE];
    char str1[MAXSIZE];
    /*
    void init(char *store) {
    int i;
    store['A']=’V', store['B']=’W',store['C']=’X',store['D']=’Y',store['E']=’Z';
    for(i=’F';i<=’Z';++i) store =i-5;
    }
    */
    int main() {
    //freopen("input.txt","r",stdin);
    //init(store);
    char *p;
    while(fgets(str1,MAXSIZE,stdin) && strcmp(str1,"STARTn")==0) {
    if(p=fgets(str1,MAXSIZE,stdin)) {
    for(;*p;++p) {
    //*p=store[*p]
    if(*p<’A’ || *p>’Z') continue;
    if(*p>’E') *p=*p-5;
    else *p=*p+21;
    }
    printf("%s",str1);
    }
    fgets(str1,MAXSIZE,stdin);
    }
    return 0;
    }