ZOJ 1006 Do the Untwish

Posted 2020-11-02 jinjin-2018

Do the Untwish

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1006

题意：给定密文按公式解密

注意点：pcode = (ccode + i)%28;的使用

贴代码：

 1 //Problem Name: Do the Untwish
 2 //Source: ZOJ 1006
 3 //Author: jinjin18
 4 //Main idea: easy to solve
 5 //Language: C++
 6 //======================================================================
 7 #include<stdio.h>
 8 #include<string.h>
 9 #include<map>
10 using namespace std;
11 
12 map<char,int> mp;
13 
14 
15 void init(){
16     mp[‘_‘] = 0;
17     mp[‘.‘] = 27;
18     for(int i = 97; i < 97+26; i++){
19         mp[i] = i - 96;
20     }
21     return;
22 }
23 
24 char Findchar(int v){
25     if(v==0){
26         return ‘_‘;
27     }
28     if(v==27){
29         return ‘.‘;
30     }
31     if(v<27&&v>0){
32         return v + 96;
33     }
34     return ‘\0‘;
35 }
36 int main(){
37 
38     init();
39     int k;
40     char ptext[100];
41     char ctext[100];
42     while(scanf("%d",&k)!=EOF && k !=0 ){
43         scanf("%s",ctext);
44         int n = strlen(ctext);
45         ptext[n] = ‘\0‘;
46         for(int i = 0; i < n; i++){
47             int ccode = mp[ctext[i]];
48             //printf("%d ",ccode);
49             //int pcode = ccode < 28-i? ccode+i:ccode - 28 + i;
50             int pcode = (ccode + i)%28;   //写成上面那行i超过28时会出错
51             //printf("%d\n",pcode);
52             ptext[(k*i)%n] = Findchar(pcode);
53 
54 
55         }
56         printf("%s\n",ptext);
57     }
58     return 0;
59 
60 }