Football Games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 802    Accepted Submission(s): 309 Problem Description A mysterious country will hold a football world championships---Abnormal Cup…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 681    Accepted Submission(s): 240 Problem Description This is a simple problem. The teacher gives Bob a list of prob…

2016 ACM/ICPC Asia Regional Qingdao Online(部分题解) 5878---I Count Two Three http://acm.hdu.edu.cn/showproblem.php?pid=5878 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1287    Accepted Submissi…

摘要 本文主要列举并求解了2016 ACM/ICPC亚洲区青岛站现场赛的部分真题,着重介绍了各个题目的解题思路,结合详细的AC代码,意在熟悉青岛赛区的出题策略,以备战2018青岛站现场赛. HDU 5984 Pocky 题意 给出一根棒子(可以吃的)的长度x和切割过程中不能小于的长度d,每次随机的选取一个位置切开,吃掉左边的一半,对右边的棒子同样操作,直至剩余的长度不大于d时停止.现在给出x和d,问切割次数的数学期望是多少. 解题思路 当看到第二个样例2 1时,结果是1.693147,联想到ln…

2016 ACM ICPC Asia Region - Tehran A - Tax 题目描述:算税. solution 模拟. B - Key Maker 题目描述:给出\(n\)个序列,给定一个序列,问\(n\)个序列中有多少个序列满足对应位的值小于或等于给定序列的值. solution 模拟. C - IOI 2017 Logo 题目描述:有\(m\)件作品,\(n\)个人投票,每个人可以选三件作品,分别给\(1, 2, 3\)分,每件作品按总分排序,总分相同按得\(3\)分的数量排序,还…

hannnnah_j’s Biological Test Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 802    Accepted Submission(s): 269 Problem Description hannnnah_j is a teacher in WL High school who teaches biolog…

I Count Two Three Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 782    Accepted Submission(s): 406 Problem Description I will show you the most popular board game in the Shanghai Ingress Resis…

QSC and Master Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 859    Accepted Submission(s): 325 Problem Description Every school has some legends, Northeastern University is the same. Enter…

odd-even number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 388    Accepted Submission(s): 212 Problem Description For a number,if the length of continuous odd digits is even and the length…

Football Games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 439    Accepted Submission(s): 157 Problem Description A mysterious country will hold a football world championships---Abnormal Cup…

Friends and Enemies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 291    Accepted Submission(s): 160 Problem Description On an isolated island, lived some dwarves. A king (not a dwarf) ruled t…

Barricade Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 997    Accepted Submission(s): 306 Problem Description The empire is under attack again. The general of empire is planning to defend his…

Function Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 976    Accepted Submission(s): 375 Problem Description The shorter, the simpler. With this problem, you should be convinced of this tru…

Sparse Graph Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 689    Accepted Submission(s): 238 Problem Description In graph theory, the complement of a graph G is a graph H on the same vertic…

Different Circle Permutation Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 208    Accepted Submission(s): 101 Problem Description You may not know this but it's a fact that Xinghai Square is…

题目链接: http://codeforces.com/gym/100851 题目大意: N个人,每个人有pi个物品,每个物品价值为0~49.每次从1~n顺序选当前这个人的物品,如果这个物品的价值>=之前所有物品价值和则加上这个物品,否则这个物品舍弃不计算在内. 总共拿出K个物品,如果一个人没物品拿了那么他会拿出价值为50的物品.求最终物品价值和有多少. 题目思路: [模拟] 直接暴力枚举.判断是否超过之前的总和,如果有人拿了50则后面的人肯定都是拿50. // //by coolxxx //#…

题目链接: http://codeforces.com/gym/100851 题目大意: n个序列.每个序列有4个值x,a,b,c,之后按照x=(a*x+b)%c扩展无穷项. 求每个序列各取一个数之后求和不是K的倍数的最大值. (x,a,b,c<=1000,n<=10000,K<=109) 题目思路: [模拟] 先暴力把每个序列能够获得的值都求出来.存下最大的两个%K不相等的值. 接下来先取每个序列最大的值,如果%K不为0则为答案. 否则把其中一个换成次优值.因为前面满足%K不相等所以只…

题目链接: http://codeforces.com/gym/100851 题目大意: 系统里生成一个字符串C,一开始告诉你字符串的长度N(偶数).接着你需要在n+500次内猜出这个字符串是什么. 每次你可以输出一个长度为N的字符串S,系统根据你输出的字符串S和C的匹配数量输入一个数,若为n/2则输入n/2,若为n则输出n,否则输入0. 只要一猜对字符串就要结束程序(即输入的数为n).如果询问次数超过n+500则出错.每次输出完要fflush(stdout). 题目思路: [模拟] 这题真的很…

题目链接: http://codeforces.com/gym/100851 题目大意: 一个N*N的矩阵A,Ai,j=i+j,Q次操作,每次分两种,R r取出第r行还未被取的所有数,并输出和.C c取出第c列还未被取出的所有数并输出和. 题目思路: [模拟] 首先Ai,j=i+j这个很关键.预处理每一行(=列)的值. 只要记录当前取得时候前面已经取过的所有行数的和.次数,所有列数的和.次数,就能推算出这次取数会少掉多少值. 并记录这一行或这一列被取过没有. // //by coolxxx //…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=5979 按AC顺序: I - Convex Time limit    1000 ms Memory limit 65536 kB OS Windows We have a special convex that all points have the same distance to origin point. As you know we can get N segments after linki…

[ HDU 5878 ] I Count Two Three 考虑极端,1e9就是2的30次方,3的17次方,5的12次方,7的10次方. 而且,不超过1e9的乘积不过5000多个,于是预处理出来,然后每次二分找就可以了. /* TASK:I Count Two Three 2^a*3^b*5^c*7^d的最小的大于等于n的数是多少 LANG:C++ URL:http://acm.hdu.edu.cn/showproblem.php?pid=5878 */ #include <iostream>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5878 题目大意: 给出一个数n ,求一个数X, X>=n. X 满足一个条件 X= 2^a*3^b*5^c*7^d 求靠近n的X值. 解题思路: 打表+二分查找 [切记用 cin cout,都是泪...] AC Code: #include<bits/stdc++.h> using namespace std; ]; int main() { ; long long ans,n,x; ; i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5884 题目大意:有n个有序的序列,对于第i个序列有ai个元素. 现在有一个程序每次能够归并k个序列, 他的花费为k个序列中的总的元素数.现在想知道在花费不超过t的情况下存在的最小的k为多少? 解题思路:二分k的值,合并k个序列的值, 加入队列,然后用哈夫曼检查最小花费是否超过t! 这时需要注意最后一次合并是否为k个数, 如果不是k个数,就不是最优的哈夫曼的值, 如果最后是k个数, 那么最后n, k一…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5889 解法:http://blog.csdn.net/u013532224/article/details/46992973 然后改改模版 #include <iostream> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <str…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5883 解法:先判断是不是欧拉路,然后枚举 #pragma comment(linker, "/STACK:102400000,102400000") #include <math.h> #include <time.h> #include <stdio.h> #include <string.h> #include <stdlib.h>…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5882 解法:一个点必须出度和入度相同就满足题意,所以加上本身就是判断奇偶性 #include<stdio.h> #include<math.h> #include<string.h> #include<stack> #include<set> #include<queue> #include<vector> #include<…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5879 解法:我们知道到某个极限之后结果相同,所以找到那个极限,其他保存之后输出就好了 #include<stdio.h> //#include<bits/stdc++.h> #include<string.h> #include<iostream> #include<math.h> #include<sstream> #include<…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5878 解法:先保存,再二分查询~具体http://blog.csdn.net/coder_xia/article/details/6707600 #include<stdio.h> //#include<bits/stdc++.h> #include<string.h> #include<iostream> #include<math.h> #inclu…

I:QSC and Master 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5900 题意: 给出n对数keyi,vali表示当前这对数的键值和权值,可以操作将连续的两个数合并,如果满足gcd(a[i],a[i+1])>1,得到的价值是两个数的权值和,每次合并两个数之后,这两个数就会消失,然后旁边的数会接上比如1 2 3 4 合并了 2 3 则 剩下1 4也可以合并 思路:区间dp 1:处理出任意区间内的所有数是否可以合并 对于当前的[l,r]…

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Rock-paper-scissors is a zero-sum hand game usually played between two people, in which each pl…