我们考虑前缀和sum[i],如果将a[i+1]中的一个塞入a[i]中,则不影响sum[i+1],但是sum[i]++,如果将a[i]中的一个塞入a[i+1],则不影响sum[i+1],但是sum[i]--,我们可以发现操作一次相当于将一个sum[i]+1或者到sum[i]-1,那么题意就变成了操作多少次可以使得所有的sum[i]为某一个k的倍数,那么一个sum[i]变成k的倍数操作次数最小必然为min(sum[i]%k,k-sum[i]%k),那么接下如何考虑k呢,显然k是sum[n]的一个因子

描述 对于一个长度为n的数列A,我们如下定义A的中位数med(A): 当n是奇数时,A的中位数是第(n+1)/2大的数:当n是偶数时,A的中位数是第n/2大的数和第n/2+1大的数的平均值. 同时,我们如下定义A的前缀中位数和: S(A) = med(B1) + med(B2) + med(B3) + ... + med(Bn) 其中Bi是A长度为i的前缀,即Bi=(A1, A2, A3 ... Ai-1, Ai),1 ≤ i ≤ n. 现在给定一个长度为n的数列X,我们希望通过将X中的数重新排

B. Alyona and a tree 题目连接: http://codeforces.com/contest/739/problem/B Description Alyona has a tree with n vertices. The root of the tree is the vertex 1. In each vertex Alyona wrote an positive integer, in the vertex i she wrote ai. Moreover, the g

题目链接: http://codeforces.com/contest/740/problem/D D. Alyona and a tree time limit per test2 secondsmemory limit per test256 megabytes 问题描述 Alyona has a tree with n vertices. The root of the tree is the vertex 1. In each vertex Alyona wrote an positiv

A. Nano alarm-clocks Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100637/problem/A Description An old watchmaker has n stopped nano alarm-clocks numbered with integers from 1 to n. Nano alarm-clocks count time in hours, and

题目链接:http://codeforces.com/contest/731/problem/F 题意:有n个数,从里面选出来一个作为第一个,然后剩下的数要满足是这个数的倍数,如果不是,只能减小为他的倍数,否则就舍弃掉,然后把没有舍弃的数的值加起来,求和的最大值; 43 2 15 9 就拿这个来说,当拿3当做第一个数时结果是3+15+9=27因为2不是3的倍数:当拿2作为第一个数时,结果是2+2+14+8=26因为3,15,9都不是2的倍数,所以只能减小;同理...求最大的和; 我们可以记录每个

题目链接:http://codeforces.com/contest/479/problem/E 题意:         给定一个启示的楼层a,有一个不能去的楼层b,对于你可以去的下一个楼层必须满足你当前楼层x与下一个要去的楼层y的距离小于x到b的距离.求出走k趟的方案数. 题解: dp[i][j]  表示第i趟 在第j层楼的方案数.一般用三个for才可以,所以我们用前缀和优化一下,时间复杂度降到O(n*k).空间复杂度有点大,所以我们可以用滚动数组. //#pragma comment(lin

题目链接:http://codeforces.com/contest/712/problem/D A初始有一个分数a,B初始有一个分数b,有t轮比赛,每次比赛都可以取[-k, k]之间的数,问你最后A比B大的情况有多少种. dpA[i][j]表示第i轮获得j分的情况数.因为第i轮只和第i-1轮有关,所以这里i用滚动数组优化. 要是普通做法3个for就会超时,所以要用前缀和优化,dpA[i][j]可以由一段连续的dp[i - 1][x]转移过来,所以用sumA数组存取dp[i - 1][x]的前缀

题目链接:http://codeforces.com/contest/703/problem/D 给你n个数,m次查询,每次查询问你l到r之间出现偶数次的数字xor和是多少. 我们可以先预处理前缀和Xor[i],表示1~i的xor和.因为num^num=0,所以Xor[r] ^ Xor[l - 1]求的是l~r之间出现奇数次的数字xor和. 那怎么求偶数次的呢,那我们可以先求l到r之间不重复出现数字的xor(比如1 1 2 求的是1 ^ 2),然后再xor以上求出的Xor[r] ^ Xor[l

C. New Year and Domino 题目连接: http://www.codeforces.com/contest/611/problem/C Description They say "years are like dominoes, tumbling one after the other". But would a year fit into a grid? I don't think so. Limak is a little polar bear who loves

B. Hamming Distance Sum 题目连接: http://www.codeforces.com/contest/608/problem/A Description Genos needs your help. He was asked to solve the following programming problem by Saitama: The length of some string s is denoted |s|. The Hamming distance betw

C. Stripe Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/18/C Description Once Bob took a paper stripe of n squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly neg

题目链接: http://codeforces.com/contest/660/problem/C 题意: 将最多k个0变成1,使得连续的1的个数最大 题解: 二分连续的1的个数x.用前缀和判断区间[i,i+x-1]里面0的个数是否小于等于k. 代码: #include<iostream> #include<cstdio> #include<vector> #include<cstring> using namespace std; ; int n,k; i

C. Riding in a Lift Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/C Description Imagine that you are in a building that has exactly n floors. You can move between the floors in a lift. Let's number the floors f

A. Nano alarm-clocks Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100637/problem/A Description An old watchmaker has n stopped nano alarm-clocks numbered with integers from 1 to n. Nano alarm-clocks count time in hours, and

B. Laurenty and Shop Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/586/problem/B Description A little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants to make sa

http://codeforces.com/contest/570/problem/D 给一棵有根树(50w个点)(指定根是1号节点),每个点上有一个小写字母,然后有最多50w个询问,每个询问给出x和f,表示询问以x为根的子树,在第f层的所有节点上的字符能否组成一个回文串 首先树形转线性,把每个点按照DFS序重新标号,然后开个vector记下第i层都有哪些节点, 对于这一层的节点,维护一个前缀和,即某个字母出现过多少次, 这样对于某个询问x,f,我们能知道x为根的子树在线性数组中的序号范围, 然

CodeForces 816B Karen and Coffee(前缀和,大量查询) Description Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally ac

题目:http://codeforces.com/contest/1154/problem/F 题意:给你n个商品,然后还有m个特价活动,你买满x件就把你当前的x件中最便宜的y件价格免费,问你买k件花最少的钱是多少 思路:首先这个特价活动就像点外卖一样满多少减多少,你可以分几次来使用同一个优惠或者不同的优惠,然后我们我们分析三个点 1,那个特价活动中x>k的我们肯定都不会使用,因为那个时候已经买满了k件 2,我们肯定是买最便宜的k件来使用特价活动 3,我们其实可以当成分开几组来使用优惠,就像外卖

It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts k days! When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically,

CodeForces - 617E 给n个数, m个询问, 每次询问问你[l, r]区间内有多少对(i, j), 使得a[i]^a[i+1]^......^a[j]结果为k.(注意 i ! =  j)维护一个前缀异或值就可以了.要注意的是 区间[l, r], 我们需要将pre[l-1]......pre[r]都加进去, pre[l-1]不能少. #include<bits/stdc++.h> using namespace std; #define maxn 1234567 #define l