ZOJ1006 Do the Untwist

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简单模拟~ #include<bits/stdc++.h> using namespace std; ; int a[maxn]; unordered_map<char,int> pos; string s="_abcdefghijklmnopqrstuvwxyz."; int len; char ch[maxn]; int main () { ;i<;i++)pos[s[i]]=i; int n; while (scanf("%d",…

1006 Do the Untwist

考察编程基础知识,用到字符和数字相互转化等.形式是描述清楚明文和暗文的转化规则. #include <stdio.h> #include <string.h> #define MAXLEN 71 int length; void toCode(int code[],char text[]){ int i; ;i<length;i++){ if(text[i] == '_') code[i] = ; else if(text[i] == '.') code[i] = ; els…

ZOJ 1006：Do the Untwist（模拟）

Do the Untwist Time Limit: 2 Seconds Memory Limit: 65536 KB Cryptography deals with methods of secret communication that transform a message (the plaintext) into a disguised form (the ciphertext) so that no one seeing the ciphertext will be able…

Do the Untwist（模拟）

ZOJ Problem Set - 1006 Do the Untwist Time Limit: 2 Seconds Memory Limit: 65536 KB Cryptography deals with methods of secret communication that transform a message (the plaintext) into a disguised form (the ciphertext) so that no one seeing the…

Do the Untwist

Do the Untwist Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 982 Accepted Submission(s): 638 Problem Description Cryptography deals with methods of secret communication that transform a…

ZOJ Problem Set - 1006 Do the Untwist

今天在ZOJ上做了道很简单的题目是关于加密解密问题的,此题的关键点就在于求余的逆运算: 比如假设都是正整数 A=(B-C)%D 则 B - C = D*n + A 其中 A < D 移项 B = A+C + D*n 当B<D时,两边对D取摸, B = B%D = ( A+C + D*n )%D = (A+C)%D 由此可得此题答案,见代码 #include <cstdio> #include <cstring> int main() { ]; ],ctext[]; w…

[ZOJ 1006] Do the Untwist (模拟实现解密)

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=6 题目大意:给你加密方式,请你求出解密. 直接逆运算搞,用到同余定理 #include <cstdio> #include <cstdlib> #include <string> #include <iostream> #include <cstring> #include <algorithm> #inc…

ACM/ICPC ZOJ1006-Do the Untwist 解题代码

#include <iostream> #include <string> #include <stdlib.h> using namespace std; int main() { string array[30][2]; string a,b; int t=0; while( cin >> a ) { if ( a == "0" ) { break; } else { cin >> b; } array[t][0] = a…

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