Instruction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 991    Accepted Submission(s): 244 Problem Description Nowadays, Jim Green has produced a kind of computer called JG. In his computer,…

http://acm.hdu.edu.cn/showproblem.php?pid=5083 机器码和操作互相转化 注意SET还要判断末5位不为0输出Error #pragma comment(linker, "/STACK:36777216") #pragma GCC optimize ("O2") #include <cstdio> #include <cstdlib> #include <cmath> #include &l…

题意:给出汇编指令,解释出编码或者给出编码,解释出汇编指令. 解法:简单模拟,按照给出的规则一步一步来就好了,主要是注意“SET”的情况,还有要输出的东西最好放到最后一起输出,中间如果一旦不对就可以及时跳出去. 其他也没什么了,只要细心点,多测几组样例就好了. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath&g…

Instruction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 347    Accepted Submission(s): 101 Problem Description Nowadays, Jim Green has produced a kind of computer called JG. In his computer…

题意: 二进制指令转汇编指令,汇编指令转二进制指令. 思路: 额,条理分好,想全,思维不能乱. 代码: int findyu(char yu[50],char c){ int l=strlen(yu); rep(i,0,l-1) if(c==yu[i]) return i; } int calc(char t[50],int x,int k){ int res=0; rep(i,x,x+k-1) res*=10, res+=(t[i]-'0'); return res; } int calc2(…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5083 题目意思:如果给出 instruction 就需要输出对应的 16-bit binary code,给出16-bit binary code 就需要输出对应的instruction. 由于不会截取的技巧,代码量非常可观 = =,所以说,一直很讨厌做模拟题!!! 留下这代码,纪念一个代码还是不够精简的自己!!!内存和时间还能接受,也比较容易理解,不过好多重复代码= =.以下这个代码可以忽略,改到…

http://acm.hdu.edu.cn/showproblem.php?pid=4891 给出一个文本,问说有多少种理解方式. 1. $$中间的,(s1+1) * (s2+1) * ...*(sn+1), si表示连续的空格数. 2.{}中间,即 | 的个数+1. 就是模拟 #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <…

http://acm.hdu.edu.cn/showproblem.php?pid=5012 模拟出骰子四种反转方式,bfs,最多不会走超过6步 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <queue> #include <vector> #include <i…

思路: 主要就是模拟这些操作,用链表果断超时.改用堆栈模拟就过了 #include<map> #include<set> #include<stack> #include<cmath> #include<queue> #include<cstdio> #include<vector> #include<string> #include<cstdlib> #include<cstring>…

The Donkey of Gui Zhou Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 389    Accepted Submission(s): 153 Problem Description There was no donkey in the province of Gui Zhou, China. A trouble m…