There are an equation. ∑0≤k1,k2,⋯km≤n∏1⩽j<m(kj+1kj)%1000000007=?∑0≤k1,k2,⋯km≤n∏1⩽j<m(kj+1kj)%1000000007=? We define that (kj+1kj)=kj+1!kj!(kj+1−kj)!(kj+1kj)=kj+1!kj!(kj+1−kj)! . And (kj+1kj)=0(kj+1kj)=0 while kj+1<kjkj+1<kj. You have to get th…

HDU 5793 - A Boring Question题意: 计算 ( ∑(0≤K1,K2...Km≤n )∏(1≤j<m) C[Kj, Kj+1]  ) % 1000000007=? (C[Kj, Kj+1] 为组合数)    分析: 利用二项式展开: (a + b) ^ n =  ∑(r = 0, n) (C[n, r] * a^(n-r) * b^r )        化简:           ∑(0≤K1,K2...Km≤n )∏(1≤j<m) C[Kj, Kj+1]       …

A Boring Question Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 156    Accepted Submission(s): 72 Problem Description       Input   The first line of the input contains the only integer T,(1≤T…

A Boring Question Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 487    Accepted Submission(s): 271 Problem Description There are an equation.∑0≤k1,k2,⋯km≤n∏1⩽j<m(kj+1kj)%1000000007=? We define…

A Boring Question 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5793 Description Input The first line of the input contains the only integer T, Then T lines follow,the i-th line contains two integers n,m. Output For each n and m,output the answer i…

A Boring Question Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 865    Accepted Submission(s): 534 Problem Description There are an equation.∑0≤k1,k2,⋯km≤n∏1⩽j<m(kj+1kj)%1000000007=?We define…

参考博客:http://www.cnblogs.com/Sunshine-tcf/p/5737627.html. 说实话,官方博客的推导公式看不懂...只能按照别人一样打表找规律了...但是打表以后其实也不是很好看出规律的...而且这个表都写了半天233...(真是太弱了= =)为了打表,我们应当先知道k数列必须是不递减的才能满足值不为0,因此我们可以用递归来写这个表(类似于dfs). AC代码如下: #include <stdio.h> #include <algorithm>…

There are an equation. ∑0≤k1,k2,⋯km≤n∏1⩽j<m(kj+1kj)%1000000007=? We define that (kj+1kj)=kj+1!kj!(kj+1−kj)! . And (kj+1kj)=0 while kj+1<kj. You have to get the answer for each n and m that given to you. For example,if n=1,m=3, When k1=0,k2=0,k3=0,(k…

题目地址:pid=4920">HDU 4920 对这个题简直无语到极点. . .竟然O(n^3)的复杂度能过....方法有三.. 1:进行输入优化和输出优化. . (前提是你的输入优化不能太搓...) 2:利用缓存优化..详情请看该论文.大体就是将后两个for循环换过来,让坐标改变的频率降下来. 3:叉姐题解中说的正规方法. .利用biset存储,进行预处理. .(事实上我还没看懂.. 我仅仅写了个另外一种.. .代码例如以下.共勉..奇妙的小代码.. #include <iostr…

题解:http://bestcoder.hdu.edu.cn/blog/ 多校6 HDU5793 A Boring Question // #pragma comment(linker, "/STACK:102c000000,102c000000") // #include <conio.h> #include <bits/stdc++.h> using namespace std; #define clc(a,b) memset(a,b,sizeof(a))…