A. Short Program link http://codeforces.com/contest/878/problem/A describe Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well. In the language…

C. Short Program time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and retur…

传送门 题目大意: 输入给出一串位运算,输出一个步数小于等于5的方案,正确即可,不唯一. 题目分析: 英文题的理解真的是各种误差,从头到尾都以为解是唯一的. 根据位运算的性质可以知道: 一连串的位运算最终都可以用三个位运算代替(&|^). 那么仅需对每一位的情况进行讨论,某一位: 必须变为1 (|1) 必须变为0 (&0) 必须01颠倒(^1) 最后输出3种位运算即可(输出三种最稳妥). code #include<bits/stdc++.h> using namespace…

题目链接:http://codeforces.com/contest/879/problem/C C. Short Program time limit per test2 seconds memory limit per test256 megabytes Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and re…

Codeforces Round #443 (Div. 2) codeforces 879 A. Borya's Diagnosis[水题] #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main(){ , s, d; scanf("%d", &n); while(n--) { scanf("%d%d", &a…

C. Short Program Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well. In the language, there are only three commands: apply a bitwise operation…

C. Short Program time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and retur…

[链接] 我是链接,点我呀:) [题意] 给你一个n行的只和位运算有关的程序. 让你写一个不超过5行的等价程序. 使得对于每个输入,它们的输出都是一样的. [题解] 先假设x=1023,y=0; 即每位都是1和每位都是0; 然后做一下这n个操作. 得出,每一位如果是0的话输出应该是几,以及每一位是1的话输出应该是几. 会发现,用or和xor和and就能完成. [代码] #include <bits/stdc++.h> #define ll long long using namespace s…

题目链接:http://codeforces.com/contest/283/problem/B 思路: dp[now][flag]表示现在在位置now,flag表示是接下来要做的步骤,然后根据题意记忆化搜索记忆,vis数组标记那些已经访问过的状态. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define REP(i, a, b) for (i…

A:考虑每一位的改变情况,分为强制变为1.强制变为0.不变.反转四种,得到这个之后and一发or一发xor一发就行了. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N…