https://nanti.jisuanke.com/t/31450 题意 给出一个映射(左为ascll值),然后给出一个16进制的数,要求先将16进制转化为2进制然后每9位判断,若前8位有奇数个1且第9位为0则这个子串取,若前8位有偶数个1且第9 位为1也取.取出的串在映射中进行查找,输出对应ascll值的字符 分析 用map直接模拟,细节需要注意. #include <bits/stdc++.h> using namespace std; #define ms(a, b) memset(a…

题意:https://nanti.jisuanke.com/t/31450 题解:题目很长的模拟,有点uva的感觉 分成四步 part1 16进制转为二进制string 用bitset的to_string() part2 parity check 校对,将处理结果pushback到另一个string part3 建字典树,用形如线段树的数组存 part4 遍历字典树 1A 233 #include<bitset> #include <cstdio> #include <cma…

Lattice's basics in digital electronics 44.08% 1000ms 131072K   LATTICE is learning Digital Electronic Technology. He is talented, so he understood all those pieces of knowledge in 10^{-9}10−9 second. In the next 10^{-9}10−9 second, he built a data d…

Lattice's basics in digital electronics 题目链接 题目描述 LATTICE is learning Digital Electronic Technology. He is talented, so he understood all those pieces of knowledge in 10−9 second. In the next 10−9 second, he built a data decoding device that decodes…

[链接] 我是链接,点我呀:) [题意] [题解] 每个单词的前缀都不同. 不能更明示了... 裸的字典树. 模拟一下.输出一下就ojbk了. [代码] #include <bits/stdc++.h> #define LL long long #define rep1(i,a,b) for (int i = a;i <= b;i++) #define rep2(i,a,b) for (int i = a;i >= b;i--) #define all(x) x.begin(),x…

原题链接:https://nanti.jisuanke.com/t/31450 附上队友代码:(感谢队友带飞) #include <bits/stdc++.h> using namespace std; #define ll long long #define FOR(i,a,b) for(int i=(a);i<=(b);++i) #define DOR(i,a,b) for(int i=(a);i>=(b);--i) const int maxN=2e5+5,inf=0x3f3…

2018 ICPC 沈阳网络赛 Call of Accepted 题目描述:求一个算式的最大值与最小值. solution 按普通算式计算方法做,只不过要同时记住最大值和最小值而已. Convex Hull 题目描述:定义函数\(gay(x)\),若\(x\)是某个非\(1\)的数的平方的倍数,则\(gay(x)=0\),否则\(gay(x)=x^2\),求\(\sum_{num=1}^{n} ( \sum_{i=1}^{num} gay(x) ) mod p\) solution \[\sum…

"Oh, There is a bipartite graph.""Make it Fantastic." X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up sev…

42.93% 1000ms 131072K LATTICE is learning Digital Electronic Technology. He is talented, so he understood all those pieces of knowledge in 10^{-9}10−9 second. In the next 10^{-9}10−9 second, he built a data decoding device that decodes data encoded w…

Given a rooted tree ( the root is node 11 ) of NN nodes. Initially, each node has zero point. Then, you need to handle QQ operations. There're two types: 1\ L\ X1 L X: Increase points by XX of all nodes whose depth equals LL ( the depth of the root i…