Progressive Scramble【模拟】

问题 J: Progressive Scramble

时间限制: 1 Sec  内存限制: 128 MB

提交: 108  解决: 45

[提交] [状态] [命题人:admin]

题目描述

You are a member of a naive spy agency. For secure communication,members of the agency use a very simple encryption algorithm – which changes each symbol in the message ‘progressively’, i.e., based on the symbols preceding it. The allowed symbols are space and the 26 lowercase English letters. For encryption purposes we assign them the values 0 (for space) and 1 through 26 (for a–z). We’ll let v(s) represent the numeric value of symbol s.

Consider a message with symbols s1, s2, . . . , sn. The encryption algorithm starts by converting the first symbol s1 into its associated value u1 = v(s1). Then for each subsequent symbol si in the message, the computed value is ui = v(si) + ui−1 — the sum of its associated value and the computed value for the previous symbol. (Note that when there is a space in the input

message, the previous scrambled letter is repeated.) This process continues until all the ui are computed.

At this point, the message is a sequence of numeric values. We need to convert it back to symbols to print it out. We do this by taking the value ui modulo 27 (since there are 27 valid symbols), and replacing that value with its corresponding symbol. For example, if ui = 32, then 32 mod 27 = 5, which is the symbol ‘e’ (since v(e) = 5).

Let’s look at an example. Suppose we want to encrypt the string “my pie”. 

1. First, convert each symbol si into v(si): [13, 25, 0, 16, 9, 5].

2. Next, compute each ui: [13, 38, 38, 54, 63, 68].

3. Then, use modulus on the ui: [13, 11, 11, 0, 9, 14].

4. Finally, convert these back to symbols: “mkk in”.

Create a program that takes text and encrypts it using this algorithm, and also decrypts text that has been encrypted with this algorithm.

输入

The input to your program consists of a single integer 1 ≤ n ≤ 100 on its own line. This number is followed by n lines, each containing the letter ‘e’ or ‘d’, a single space, and then a message made up of lowercase letters (a–z) and spaces, continuing to the end of the line. Each message is between 1 and 80 characters long. The letters ‘d’ and ‘e’ indicate that your program decrypts or encrypts the subsequent string, respectively.

输出

Output the result of encrypting or decrypting each message from the input on its own separate line. Note that differences in whitespace are significant in this problem. Therefore your output must match the correct output character-for-character, including spaces.

样例输入

7
e testing multiple letters rrrrrrrrrrrrr
e this particularly long  sentence can test encryption
d tajbbrsjcloiuvmywwhwjqqqinauzmpuuxyllejbvv nqhfvoxlz
e my pie
d mkk in
e the quick brown fox jumps over the lazy dog
d taffwqzbmmofuqddjyvvezlatthchzzs eeqrqoosgn

样例输出

tyqjsfmmzteygwhmmycwpulddvmdvmdvmdvmdv
tajbbrsjcloiuvmywwhwjqqqinauzmpuuxyllejbvv nqhfvoxlz
this particularly long  sentence can test encryption
mkk in
my pie
taffwqzbmmofuqddjyvvezlatthchzzs eeqrqoosgn
the quick brown fox jumps over the lazy dog

很水的一个模拟阿，但比赛时候一直狂怼格式错误，然后瞎改改成WA？ 赛后zn说漏掉了字符串第一个就是空格的情况=。=

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define rep(i,a,n) for(int i=a;i<n;++i)
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define sca(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define mst(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
typedef pair<int,int> P;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int inf =0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 105;
const int maxn = 10010;

int n, t;
int ans[maxn];
string s;
int main()
{
  read(t);
  getchar();
  while(t--){
    getline(cin,s);
    ans[2] = s[2] == ' ' ? 0 : s[2] - 'a' + 1;
    if(s[0] == 'e'){
      for(int i = 3; i < s.length(); i++){
        int x = s[i] == ' ' ? 0 : s[i] -'a' + 1;
        ans[i] = ans[i - 1] + x;
      }
      for(int i = 2; i < s.length(); i++){
        printf("%c",ans[i] % 27 ? ans[i] % 27 - 1 + 'a' : ' ');
      }
      printf("\n");
    }
    else{
      int sum = ans[2];
      for(int i = 3; i < s.length(); i++){
        int x = s[i] == ' ' ? 0 : s[i] -'a' + 1;
        while(x < sum)
          x += 27;
        ans[i] = x - sum;
        sum += ans[i];
      }
      for(int i = 2; i < s.length(); i++){
        printf("%c",ans[i] % 27 ? ans[i] % 27 - 1 + 'a' : ' ');
      }
      printf("\n");
    }
  }
  return 0;
}