传送门 题意 给出n个数及x,求 \[\frac{\sum _{i=1}^n x^{a_1+a_2+...+a_{i-1}+a_{i+1}+...a_n}}{\prod_{i=1}^n x^{a_i}}\] 分析 结果必然为\(x^{sum}\),sum的值首先取所有数的和减去最大值 然后暴力合并,具体原因我不太懂,只能附上CF的标准题解 Obviously, the answer is \(x^v\). Let \(sum = a1 + a2 + ... + an\). Also let $s…

题目传送门 /* 二分查找/暴力:先埃氏筛选预处理,然后暴力对于每一行每一列的不是素数的二分查找最近的素数,更新最小值 */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; ; const int INF = 0x3f3f3f3f; int a[MAXN][MAXN]; int mn_r[MAXN]; int mn_c[MAXN]; bool is_prim…

A. Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Simon has a rectangular table consisting of n rows and m columns. Simon numbered the rows of the table from top to bottom starting f…

引子: A题过于简单导致不敢提交,拖拖拉拉10多分钟还是决定交,太冲动交错了CE一发,我就知道又要错过一次涨分的机会.... B题还是过了,根据题意目测数组大小开1e5,居然蒙对,感觉用vector更好一点... C题WA第9组,GG思密达....明天起床再补C吧 - - 题目链接: http://codeforces.com/contest/835/problem/B B. The number on the board Some natural number was written on t…

D. Pair of Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l…

链接: https://codeforces.com/contest/1215/problem/B 题意: You are given a sequence a1,a2,-,an consisting of n non-zero integers (i.e. ai≠0). You have to calculate two following values: the number of pairs of indices (l,r) (l≤r) such that al⋅al+1-ar−1⋅ar…

D. Minimum number of steps time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of s…

E - Bus Number 最近感觉打CF各种车祸.....感觉要反思一下, 上次读错题,这次想当然地以为18!肯定暴了longlong 而没有去实践, 这个题我看到就感觉是枚举每个数字的个数,但是我觉得算得时候会爆longlong 就想用大数,但是我去看别人交的全部都是C++,就感觉是不是有别的方法, 想了半天感觉时间来不及了就强行上了个java,结果时间来不及... 以后写题首先要读清楚题目,其次不要想当然,要去实践!!!!!!!!!!! 真的很烦. import java.math.Bi…

题目链接:http://codeforces.com/contest/432/problem/C 首先由题意分析出:这些数是从1到n且各不相同,所以最后结果肯定是第i位的数就是i. 采用这样一种贪心策略:从1到n枚举每一位.如果第i位不是i,那么就把i 从它所在的位置移动到第i 位,每次移动的距离要选取符合要求的最大的素数,那么由哥德巴赫猜想可以知道,最后肯定小于5次移动就移好了.每一位都这么移就行了.这过程中记录一下每次谁和谁交换位置就行了. 我把x[],y[]开成了maxn而导致了RE,改成…

B. Perfect Number time limit per test2 seconds memory limit per test256 megabytes Problem Description We consider a positive integer perfect, if and only if the sum of its digits is exactly 10. Given a positive integer k, your task is to find the k-t…

链接: https://codeforces.com/contest/1271/problem/E 题意: At first, let's define function f(x) as follows: f(x)={x2x−1if x is evenotherwise We can see that if we choose some value v and will apply function f to it, then apply f to f(v), and so on, we'll…

题目传送门 B. Perfect Number time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output We consider a positive integer perfect, if and only if the sum of its digits is exactly 1010. Given a positive integ…

B. Long Number time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a long decimal number aa consisting of nn digits from 11 to 99. You also have a function ff that maps every dig…

解题思路: 如果序列a是单调递增的,则序列为1,2,..... 2n,则将给出的式子化简得Σ(a2i - a2i-1) = n 如果序列a是单调递减的,则序列为2n,.........2, 1,则将给出的式子化简得为0 故本题的解决方法是前面2k个序列即a1,a2......a2k是单调递增的,后面的序列是单调递减的 #include <iostream> using namespace std; int main(){ int n , k; cin >> n >> k…

#include <iostream> #include <vector> using namespace std; int main(){ int n,m; cin >> n >> m; ][]; bool flag = false; ; i < n ; ++ i){ ; j < m ; ++ j){ cin >> a[i][j]; } } ; j < m; ++ j){ ][j] || a[n-][j]){ flag = t…

#include <iostream> #include <vector> using namespace std; int main(){ ; cin >> n >>k; ; i < n ; ++ i ){ int a; vector<,); cin >> a; while(a){ <=k) goodNum[a%]++; a/=; } ; ; j < k+; ++ j) if(!goodNum[j]) break; )…

刷了一页的WA  ..终于发现了 哪里错了 快速幂模板里一个变量t居然开得long  ... 虽然代码写的丑了点 但是是对的 那个该死的long 啊.. #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> using namespace std; #define mo…

A: 要么是两次要么4次,判断是否在边界: #include<cstdio> using namespace std; int main() { int n,m,x; ; scanf("%d%d",&n,&m); ; i<n; i++) ; j<m; j++) { scanf("%d",&x); &&(i==||i==n-||j==||j==m-)) flag=; } "); ");…

题目大意:给你n个点的一棵树, 每个点的权值为2^i ,让你删掉k个点使得剩下的权值和最大. 思路:这题还是比较好想的, 我们反过来考虑, 剩下一个的情况肯定是选第n个点,剩下两个 我们肯定优先考虑第n - 1 个点, 因为其他点全部加起来都没有这个点的权值大, 所以我们可以 以第n个点为根, 倍增出每个点祖先的情况, 然后从后往前贪心, 能取到就取, 不能取到就跳过. #include<bits/stdc++.h> #define LL long long #define fi first…

最后肯定是bbbb...aaaa...这样. 你每进行一系列替换操作,相当于把一个a移动到右侧. 会增加一些b的数量……然后你统计一下就行.式子很简单. 喵喵喵,我分段统计的,用了等比数列……感觉智障.一个a一个a地统计答案即可. #include<cstdio> #include<cstring> #include<iostream> using namespace std; #define MOD 1000000007ll typedef long long ll;…

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题意:对于一个数\(x\),有函数\(f(x)\),如果它是偶数,则\(x/=2\),否则\(x-=1\),不断重复这个过程,直到\(x-1\),我们记\(x\)到\(1\)的这个过程为\(path(x)\),它表示这个过程中所有\(x\)的值,现在给你一个数\(n\)和\(k\),要你找出最大的数\(x\),并且\(x\)在\(path[1,n]\)中出现的次数不小于\(k\). 题解:我们对于\(x\)可以选择分奇偶来讨论. 1.假如\(x\)为奇数,那么根据题意,我们知道\(x,2x,2x…

题目传送门 /* 题意:b+1,b+2,...,a 所有数的素数个数和 DP+埃氏筛法:dp[i] 记录i的素数个数和,若i是素数,则为1:否则它可以从一个数乘以素数递推过来 最后改为i之前所有素数个数和,那么ans = dp[a] - dp[b]: 详细解释:http://blog.csdn.net/catglory/article/details/45932593 */ #include <cstdio> #include <algorithm> #include <cs…

题目传送门 /* 题意:这题就是求b+1到a的因子个数和. 数学+DP:a[i]保存i的最小因子,dp[i] = dp[i/a[i]] +1;再来一个前缀和 */ /************************************************ Author :Running_Time Created Time :2015-8-1 14:08:34 File Name :B.cpp ************************************************…

今天老师(orz sansirowaltz)让我们做了很久之前的一场Codeforces Round #257 (Div. 1),这里给出A~C的题解,对应DIV2的C~E. A.Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular cho…

 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 #346 (Div. 2)---E. New Reform E. New Reform time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Berland has n cities connected by m bidirectional roads. No road connects a city…

Codeforces Round #372 (Div. 2) C. Plus and Square Root 题意 一个游戏中,有一个数字\(x\),当前游戏等级为\(k\),有两种操作: '+'按钮:使得\(x=x+k\) '√'按钮:使得\(x=\sqrt{x}\),此时\(x\)必须是平方数,游戏等级加1,即\(k=k+1\),且\(\sqrt{x}\)是\(k+1\)的倍数. 游戏开始时,\(x=2,k=1\),输出\(n(n \le 10^5)\)个数,表示每个等级对应的\(\frac…

A. Sereja and Dima time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja and Dima play a game. The rules of the game are very simple. The players have n cards in a row. Each card contains…

Vasiliy's Multiset 题目链接: http://codeforces.com/contest/706/problem/D Description Author has gone out of the stories about Vasiliy, so here is just a formal task description. You are given q queries and a multiset A, initially containing only integer…