题目链接 Dasha and Very Difficult Problem 求出ci的取值范围,按ci排名从小到大贪心即可. 需要注意的是,当当前的ci不满足在这个取值范围内的时候,判为无解. #include <bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i(a); i <= (b); ++i) + ; int a[N], b[N], c[N], op[N]; int l, r, L, R, cnt; b…

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Dasha logged into the system and began to solve problems. One of them is as follows: Given two sequences a and b of length n each you need t…

D. Dasha and Very Difficult Problem 题目连接: http://codeforces.com/contest/761/problem/D Description Dasha logged into the system and began to solve problems. One of them is as follows: Given two sequences a and b of length n each you need to write a se…

题目链接:http://codeforces.com/contest/761/problem/D D. Dasha and Very Difficult Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Dasha logged into the system and began to solve problem…

D. Dasha and Very Difficult Problem time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Dasha logged into the system and began to solve problems. One of them is as follows: Given two sequences…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Dasha logged into the system and began to solve problems. One of them is as follows: Given two sequences a and b of length n each you need to wr…

题目链接:http://codeforces.com/contest/761/problem/D 题意:给出一个长度为n的a序列和p序列,求任意一个b序列使得c[i]=b[i]-a[i],使得c序列的大小顺序和p序列一样. p序列给出的的是1-n不重复的数.还有给出的l和r是b序列的大小范围 显然为了要使结果存在尽量使b取得最接近r,所以可以先按p排序一下,从大到小处理b,二分l,r的值使得每次的b尽量 大. #include <iostream> #include <cstring&g…

http://codeforces.com/contest/761/problem/D c[i] = b[i] - a[i],而且b[]和a[]都属于[L, R] 现在给出a[i]原数组和c[i]的相对大小,要确定b[i] 因为已经知道了c[i]的相对大小,那么从最小的那个开始,那个肯定是选了L的了,因为这样是最小的数, 然后因为c[i]要都不同,那么记录最小的那个c[]的大小是mx,那么下一个就要是mx + 1,就是倒数第二小的那个. 也就是b[i]在[L, R]中选一个数,使得b[i] -…

Educational Codeforces Round 40 F. Runner's Problem 题意: 给一个$ 3 * m \(的矩阵,问从\)(2,1)$ 出发 走到 \((2,m)\) 的方案数 \(mod 1e9 + 7\), 走的规则和限制如下: From the cell (i, j) you may advance to: (i - 1, j + 1) - only if i > 1, (i, j + 1), or (i + 1, j + 1) - only if i <…

A Simple But Difficult Problem Time Limit: 5000ms Memory Limit: 65536KB 64-bit integer IO format: %lld      Java class name: Main Prev Submit Status Statistics Discuss Next Type: None   None   Graph Theory       2-SAT       Articulation/Bridge/Biconn…

A - Noldbach problem 题面链接 http://codeforces.com/contest/17/problem/A 题面 Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nic…

http://codeforces.com/contest/459/problem/D 题意:给你n个数,然后统计多少组(i,j)使得f(1,i,ai)>f(j,n,aj); 思路:先从左往右统计f(1,i,ai)记录在b[i],然后从右往左统计f(j,n,aj)记录在c[i],然后利用树状数组统计个数即可. #include <cstdio> #include <iostream> #include <cstring> #include <cmath>…

Pashmak and Parmida's problem Time Limit:3000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 459D Description Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she…

C. Commentator problem 题目连接: http://www.codeforces.com/contest/2/problem/C Description The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more…

题目链接  Educational Codeforces Round 40  Problem I 题意  定义两个长度相等的字符串之间的距离为:   把两个字符串中所有同一种字符变成另外一种,使得两个字符串相等所需要操作的次数的最小值.   求$s$中每一个长度为$t$的长度的连续子串与$t$的距离.字符集为小写字母$a$到$f$ 首先解决求两个长度相等的字符串之间的距离这个问题. $s$和$t$相同位上的字母连一条无向边,最后的答案是$s$和$t$中所有出现过的字符的个数减去这个无向图的连通块…

[题目链接]:http://codeforces.com/contest/798/problem/C [题意] 给你n个数字; 要求你进行若干次操作; 每次操作对第i和第i+1个位置的数字进行; 将 a[i]变为a[i]-a[i+1],a[i+1]变为a[i]+a[i+1]; 然后问你能不能通过以上变换使得最后所有元素的gcd>1 [题解] 答案总是存在的; 因为把这个操作进行两次可以得到-2*a[i+1],2*a[i] 也就是如果对每相邻的元素进行操作,最后都能使得这两个元素都变成偶数; 最后…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output There are some beautiful girls in Arpa's land as mentioned before. Once Arpa came up with an obvious problem: Given an array and a number x, coun…

题目链接:http://codeforces.com/contest/798/problem/C 题意:给出一串数字,问如果这串数字的gcd大于1,如果不是那么有这样的操作,删除ai, ai + 1 并且把 ai - a(i + 1), ai + a(i + 1) 放入原来的位置.问是否能够在几步操作后使得串的gcd大于1然后要求最小的操作数. 题解:偶数=偶数*偶数 or 奇数*偶数,奇数=奇数*奇数. 如果整个字符串全是偶数的话肯定gcd是大于1的.介于题目要求的操作,奇数-(or)+奇数=…

原题: Description Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry problem. Let's define f([l, r]) = r - l + 1 to be the number of integer points in the segment [l, r] with l ≤ r (say that ). Y…

Hard problem 题意: 有n个字符串,对第i个字符串进行反转操作代价为ci. 要使n个字符串按照字典序从小到大排列,最小的代价是多少. 题解: 反转就是reverse操作,比如说45873反转之后只能是37845,不能是别的,当时就这没有理解好,所以没继续去想,其实可以假设这样,之后来一发的,如果当时这样不就对了嘛. 一行只有反转或不反转,要求最小代价,有道01背包的意思,所以就要dp试试,dp[i]代表第i的串的最小代价,但是怎么继续推呢?,没有办法,就再多开一维,另一维只要代表2个…

https://vjudge.net/problem/CodeForces-761B 题意: 有一个圆形跑道,上面有若干个障碍,分别给出两个人距离障碍的距离,问这两个人是否是在同一个跑道上跑步(我是这么理解的,障碍的相对位置相同,那么他们就在同一个跑道上). 思路: 根据距离计算出障碍之间的距离,然后算两个跑道的距离序列是否一样,当然会有偏移的量,这个时候枚举偏移量就可以了.orz,赛上的时候就是定了一个相同位置就是偏移量在,真是sb. 代码: #include <stdio.h> #incl…

C. Little Elephant and Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The Little Elephant has got a problem — somebody has been touching his sorted by non-decreasing array a of le…

[题目]C. Sonya and Problem Wihtout a Legend [题意]给定n个数字,每次操作可以对一个数字±1,求最少操作次数使数列递增.n<=10^5. [算法]动态规划+前缀和优化 [题解]★令b[i]=a[i]-i,则a[i]递增等价于b[i]不递减. 这样做之后,显然数字加减只能到b[i]中出现的数字,而不会出现其它数字. 令f[i][j]表示前i个数,第i个数字大小为c[j](第j大的数字)的最少操作次数. f[i][j]=abs(b[i]-c[j])+min{f…

  B. Far Relative’s Problem   time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has n friends and each of th…

A. Hungry Student Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some frie…

题目链接 Dasha and Password 题目保证一定有解. 考虑到最多只有两行的指针需要移动,那么直接预处理出该行移动到字母数字或特殊符号的最小花费. 然后O(N^3)枚举求最小值即可. 时间复杂度O(N*M+N^3) #include <bits/stdc++.h> using namespace std; #define REP(i,n) for(int i(0); i < (n); ++i) #define rep(i,a,b) for(int i(a); i <=…

题目链接 Dasha and Puzzle 对于无解的情况:若存在一个点入度大于4,那么直接判断无解. 从根结点出发(假设根结点的深度为0), 深度为0的节点到深度为1的节点的这些边长度为2^30, 深度为1的节点到深度为2的节点的这些边的长度为2^29, ……………………………………………………………… 以此类推. 因为结点个数最多只有30个,所以长度分配足够. #include <bits/stdc++.h> using namespace std; #define REP(i,n) fo…

题目链接: E. Mike and Geometry Problem time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geomet…

给定一个函数: f([l,r]) = r - l + 1; f(空集) = 0; 即f函数表示闭区间[l,r]的整点的个数 现在给出n个闭区间,和一个数k 从n个区间里面拿出k个区间,然后对这k个区间求并集,并求并集的f函数值 求所有C(n,k)种方案的f函数值之和 1 <= k <= n <= 200000 -10^9 <= l <= r <= 10^9 思路: 思路其实很容易想到 对这些区间缩点 g(i) 表示i这个点代表的区间的点数(即点i实际的点数) s(i)…

题意:给定一个矩阵,表示每两个节点之间的权值距离,问是否可以对应生成一棵树, 使得这棵树中的任意两点之间的距离和矩阵中的对应两点的距离相等! 思路:我们将给定的矩阵看成是一个图,a 到 b会有多条路径, 如果存在一棵树,那么 这个树中a->b的距离一定是这个图中所有a->b中路径长度最短的一条!所以我们根据边权, 建立一棵MST树!再将MST树中的任意两点之间的距离求出来,看是否和矩阵中的对应的节点 对距离相同! #include<iostream> #include<cst…