解题思路是: Q=q1^q2.......^qn = p1^p2......^pn^((1%1)^....(1%n))^((2%1)^......(2%n))^.... 故Q的求解过程分成两部分 第一部分是求p1^p2......^pn 第二部分是求((1%1)^....(1%n))^((2%1)^......(2%n))^.... 将其化成矩形的形式 1%1   1%2  ...........  1%n 2%1   2%2  ............ 2%n ................…

题目 比赛的时候找出规律了,但是找的有点慢了,写代码的时候出了问题,也没交对,还掉分了.... 还是先总结一下位移或的性质吧: 1.  交换律 a ^ b = b ^ a 2. 结合律 (a^b) ^ c = a ^ (b^c) 3. 0^a = a; 4. a^a = 0;    a^a^a = a; 5.   知道a,b,c中任意两个就能推知第三个.      a^b = c 两边同时与a异或得: a ^ (a^b) = a^c 即 0^b = a^c  亦即 b = a^c 四个也是一样…

A. Laptops time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the…

题意: 我们在研究罗马数字.罗马数字只有4个字符,I,V,X,L分别代表1,5,10,100.一个罗马数字的值为该数字包含的字符代表数字的和,而与字符的顺序无关.例如XXXV=35,IXI=12． 现在求问一个长度为 nnn 的罗马数字可以有多少种不同的值. n<=109n<=10^9n<=109. 题解: 我们可以用暴力的方法求出前20项的值,其中前111111 项采用打表的方式,而从第12项开始答案始终 = 前一项的答案 + 49,即 292+(n−11)∗49292+(n-11)*…

// Codeforces Round #365 (Div. 2) // C - Chris and Road 二分找切点 // 题意:给你一个凸边行,凸边行有个初始的速度往左走,人有最大速度,可以停下来,竖直走. // 问走到终点的最短时间 // 思路: // 1.贪心来做 // 2.我觉的二分更直观 // 可以抽象成:一条射线与凸边行相交,判断交点.二分找切点 #include <bits/stdc++.h> using namespace std; #define LL long lon…

题目链接 A. Squats time limit per test:1 secondmemory limit per test:256 megabytesinput:standard inputoutput:standard output Pasha has many hamsters and he makes them work out. Today, n hamsters (n is even) came to work out. The hamsters lined up and eac…

CF424 A. Squats 题目意思: 有n(n为偶数)个x和X,求最少的变换次数,使得X的个数为n/2,输出变换后的序列. 解题思路: 统计X的个数ans,和n/2比較,少了的话,须要把n/2-ans个x变成X,多了的话须要把ans-n/2个X变成x.(从前往后扫一遍即可了). 代码: //#include<CSpreadSheet.h> #include<iostream> #include<cmath> #include<cstdio> #incl…

题目链接:http://codeforces.com/contest/424/problem/C, 想来一个小时,就是做不出,都做出来了,悲剧! 分析:我们知道交换异或的顺序不影响答案! 然后就是求t=a1^a2^a3^.....^an;这个直接做可以, q=(1%i)^(2%i)... ^(n^i);(1<=i<=n); //继续写完 当时我的想法是通过找规律,把1-n-1的个数都找出来,但是这个规律找不出. 后来知道有循环节这回事 比如:i=10的时候,循环节是1,2,3,4,..9,如果…

A. Magic Spheres Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606/problem/A Description Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the…

D. Magic Breeding link http://codeforces.com/contest/878/problem/D description Nikita and Sasha play a computer game where you have to breed some magical creatures. Initially, they have k creatures numbered from 1 to k. Creatures have n different cha…

D1. Magic Powder - 1 题目连接: http://www.codeforces.com/contest/670/problem/D1 Description This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single soluti…

题目链接: http://codeforces.com/contest/670/problem/D2 题解: 二分答案. #include<iostream> #include<cstdio> #include<cstring> #include<map> using namespace std; + ; const int INF = 2e9; typedef __int64 LL; int n, k; int x[maxn], y[maxn]; bool…

A. Magic Spheres   Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he nee…

D2. Magic Powder - 2 time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The term of this problem is the same as the previous one, the only exception — increased restrictions. Input The first l…

B. Magic Forest time limit per test 1 second memory limit per test 256 megabytes Problem Description Imp is in a magic forest, where xorangles grow (wut?) A xorangle of order n is such a non-degenerate triangle, that lengths of its sides are integers…

A. Magic Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of…

注意题目一次只能改变一个松鼠,Pasha can make some hamster ether sit down or stand up.是单数不是复数 #include <iostream> #include <string> #include <vector> #include <algorithm> #include <cmath> using namespace std; int main(){ int n; cin >>…

按照半径排序,然后累加人数直到超过百万 #include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <map> using namespace std; int main(){ int n,s; cin >> n >>s; map<double,int> a; ; i < n ; ++ i){ d…

#include <iostream> #include <vector> #include <algorithm> #include <string> using namespace std; int main(){ long long n; cin >>n; while(n){ if(n%10 == 1) n/=10; else if(n%100 == 14 ) n/=100; else if(n%1000 == 144) n/=1000;…

一看就是找规律的题.只要熟悉异或的性质,可以秒杀. 为了防止忘记异或的规则,可以把异或理解为半加运算:其运算法则相当于不带进位的二进制加法. 一些性质如下: 交换律: 结合律: 恒等律: 归零律: 典型应用:交换a和b的值:a=a^b^(b=a); #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<…

https://codeforces.com/contest/1131/problem/D 题意 给你n个字符串,字符串长度总和加起来不会超过1e5,定义字符串相乘为\(s*s1=s1+s[0]+s1+s[1]+s1+...+s1+s[size-1]+s1+s[size]+s1\),求n个字符串依次相乘后最长连续字符相同的子序列长度 题解 鬼畜的题意 or 难以优化的复杂度,都需要观察性质才能做,第二串要插入第一个串每个字符之间,可以看出字符数增长的速度很快,所以并不能把整个字符存下来 只看一种…

传送门 Description Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a…

E. New Reform Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connected by no more than one road. It is not guaranteed that you can get from any city to any other one, using only…

这场比赛的出题人挺有意思,全部magic成了青色. 还有题目中的图片特别有趣. 晚上没打,开virtual contest打的,就会前三道,我太菜了. 最后看着题解补了第四道. 比赛传送门 A. Angry Students 题目大意:有t队学生,每个学生有两种状态,生气(A)或不生气(P).(话说为什么生气的戴着圣诞帽哇)所有生气的人都会往前一个人丢雪球,被丢到的人也会变得生气,也会丢雪球.问你每队人中最后一个学生变得生气的时刻. 这题就是统计最长的连续的'P'当然前提是左边有生气的人. 代码…

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…

Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…