题目链接:http://poj.org/problem?id=1298 思路分析:水题,字符偏移求解,注意字符串输入问题即可. 代码如下: #include <iostream> #include <string> using namespace std; + ; char A[MAX_N]; int main() { int Len; ]; while ( scanf( "%s", Tmp )!= EOF ) { ) break; getchar( ); ge…

题目链接:http://poj.org/problem?id=1298 题目大意:按照所给的顺序要求将输入的字符串进行排列. #include <iostream> #include <cstdio> #include <cstring> using namespace std; int main () { ]= {"VWXYZABCDEFGHIJKLMNOPQRSTU"}; ]; ) { gets(ch); )//strcmp是字符的一个函数,也就…

Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever faced was keeping himself alive. In order for him to survive, he decided to create one of the first ciphers. This cipher was so incredibly sound, that no one coul…

The Hardest Problem Ever Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u   Description Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever faced was keeping himself alive. In order for…

The Hardest Problem Ever Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24039   Accepted: 13143 Description Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever faced was keeping himself alive. In orde…

The Hardest Problem Ever 链接:http://acm.hdu.edu.cn/showproblem.php?pid=1048 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 31307    Accepted Submission(s): 14491 Problem Description Julius Caesa…

The Hardest Problem Ever Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13035    Accepted Submission(s): 5905 Problem Description Julius Caesar lived in a time of danger and intrigue. The harde…

The Hardest Problem Ever Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13590    Accepted Submission(s): 6212 Problem Description Julius Caesar lived in a time of danger and intrigue. The hard…

链接:http://poj.org/problem?id=1298&lang=default&change=true 简单的入门题目也有这么强悍的技巧啊!! 书上面的代码: 很厉害有没有=_= n;main(k){,&n,);k-=n<)k%||putchar(n%<?n:n%%+);} 72字节. 用到了read这个函数 对于不需要处理的字符串,没有作处理,简化了很多.…

题目:http://poj.org/problem?id=1298 好吧,给了题目也看不懂……给出翻译(题目名翻译是:最难的问题,233333) 这一看就是老师给出题解: 然而没有什么用哈 最快的办法是,把下面的密文直接拷过来,建个字符数组 然后读入,判断是否是 ENDOFINPUT 不是就读入原文,把大写字母翻译出来 记得换行 最后读掉 END #include<cstdio> #include<cstdlib> #include<cstring> #include&…