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…

I will show you the most popular board game in the Shanghai Ingress Resistance Team. It all started several months ago. We found out the home address of the enlightened agent Icount2three and decided to draw him out. Millions of missiles were detonat…

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…

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description I will show you the most popular board game in the Shanghai Ingress Resistance Team.It all star…

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description I will show you the most popular board game in the Shanghai Ingress Resistance Team.It all star…

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…

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…

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…

[ 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=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…