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…

传送门 zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 575    Accepted Submission(s): 181 Problem Description As one of the most powerful brushes, zhx is required to give his juniors…

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 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…

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>…

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…

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…

题目链接:https://nanti.jisuanke.com/t/31716 题目大意:有n个孩子和n个糖果,现在让n个孩子排成一列,一个一个发糖果,每个孩子随机挑选x个糖果给他,x>=1,直到无糖果剩余为止.给出数字n,问有多少种分发糖果的方法. 样例输入 复制 1 4 样例输出 复制 8 解题思路:我们可以这样想,一个糖果的话,应该是只有1种方法记为x1,如果是两个糖果的话,有两种方法即为x2,分别为(1,1)和(2),从中我们可以想到如果n个糖果的话,就可以分为第n个人取1个的话就有x(…

链接:https://ac.nowcoder.com/acm/contest/153/1047 来源:牛客网 题目描述 给定一个正整数k( ≤ k ≤ ),把所有k的方幂及所有有限个互不相等的k的方幂之和构成一个递增的序列,例如,当k = 3时,这个序列是: ,,,,,,,…(该序列实际上就是:,,+,,+,+,++,…) 请你求出这个序列的第N项的值(用10进制数表示).例如,对于k = ,N = ,正确答案应该是 . 输入描述: 输入1行,为2个正整数,用一个空格隔开:k N(k.N的含义与…

题目链接: Reading comprehension Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others) Problem Description   Read the program below carefully then answer the question.#pragma comment(linker, "/STACK:1024000000,1024000000&quo…

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…

Problem Description An Arc of Dream is a curve defined by following function: where a0 = A0 ai = ai-*AX+AY b0 = B0 bi = bi-*BX+BY What ,,,?   Input There are multiple test cases. Process to the End of File. Each test nonnegative integers as follows:…

题目链接: All X Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description   F(x,m) 代表一个全是由数字x组成的m位数字.请计算,以下式子是否成立: F(x,m) mod k ≡ c   Input   第一行一个整数T,表示T组数据.每组测试数据占一行,包含四个数字x,m,k,c 1≤x≤9 1≤m≤10^10 0≤c<k≤10,…

Happy 2004 问题描述 : Consider a positive integer X,and let S be the sum of all positive integer divisors of 2004^X. Your job is to determine S modulo 29 (the rest of the division of S by 29). Take X = 1 for an example. The positive integer divisors of 2…

1.乘法逆元 直接使用等比数列求和公式,注意使用乘法逆元 ---严谨,失细节毁所有 #include "bits/stdc++.h" using namespace std; #define rep(i, s, n) for(int i=s;i<n;i++) #define MOD 1000000007 #define LL long long ; LL quick_pow(LL a,LL b) { LL ans=; ){ ){ ans=ans*a%MOD; } b>>…

普通NIM规则加上一条可以分解为两堆,标准的Multi-SG游戏 一般Multi-SG就是根据拓扑图计算SG函数,这题打表后还能发现规律 sg(1)=1 sg(2)=2 sg(3)=mex{0,1,2,1^2}=4 sg(4)=mex{0,1,2,sg(3)}=3 可以发现3和4的时候相当于互换了位置 /** @Date : 2017-10-12 21:20:21 * @FileName: HDU 3032 博弈 SG函数找规律.cpp * @Platform: Windows * @Autho…

题意: 给出n次翻转和m张牌,牌相同且一开始背面向上,输入n个数xi,表示xi张牌翻转,问最后得到的牌的情况的总数. 思路: 首先我们可以假设一开始牌背面状态为0,正面则为1,最后即是求ΣC(m,k),k为所有能取到1的情况.首先我们要确认最后1的奇偶性.因为一次翻转0->1,或者1->0,则最后所有1的情况的奇偶性相同.然后我们要找到最小的1的个数i和最大的1的个数j,i为能翻1则翻1,j为能翻0则翻0,介于中间的情况是取偶数步数,一半翻1,一半翻0,保持1的个数不变.那么k为(i<=…

题目链接 :http://acm.hdu.edu.cn/showproblem.php?pid=6030 Problem Description Little Q wants to buy a necklace for his girlfriend. Necklaces are single strings composed of multiple red and blue beads. Little Q desperately wants to impress his girlfriend,…

题目链接 Problem Description Function Fx,ysatisfies: For given integers N and M,calculate Fm,1 modulo 1e9+7. Input There is one integer T in the first line. The next T lines,each line includes two integers N and M . 1<=T<=10000,1<=N,M<2^63. Output…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1030 Delta-wave Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7163    Accepted Submission(s): 2772 Problem Description A triangle field is numbe…

已知有n个单位的水,问有几种方式把这些水喝完,每天至少喝1个单位的水,而且每天喝的水的单位为整数.看上去挺复杂要跑循环,但其实上,列举几种情况之后就会发现是找规律的题了= =都是2的n-1次方,而且这题输出二进制数就行了......那就更简单了,直接输出1,然后后面跟n-1个0就行了╮(╯_╰)╭ 下面AC代码 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm>…

Problem about GCD Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 470    Accepted Submission(s): 77 Problem Description Given integer m. Find multiplication of all 1<=a<=m such gcd(a, m)=1 (cop…

题目 //找规律,123321123321123321…发现这样排列恰好可以错开 // 其中注意题中数据范围: M是行,N是列,3 <= N < 2×M //则猜测:m,m,m-1,m-1,m-2,m-2,……,2,2,1,1求出前m个数字的和就是答案. //发现案例符合(之前的代码第二天发现案例都跑不对,真不知道我当时眼睛怎么了) #include <iostream> #include<stdio.h> #include<string.h> #inclu…

主要还是找规律,然后大数相乘 #include<stdio.h> #include<string.h> #include<math.h> #include<time.h> #include<map> #include<iostream> #include<ctype.h> #include<string> #include<algorithm> #include<stdlib.h> #i…

Starting from point (0,0) on a plane, we have written all non-negative integers 0, 1, 2,... as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3, 1) respectively and this pattern has continued. You are to w…

M=1:aaaaaaaa…… M=2:DFS+manacher, 暴出N=1~25的最优解,找规律.N<=8的时候直接输出,N>8时,头两个字母一定是aa,剩下的以aababb循环,最后剩余<5全部补a,等于5补aabab. M=3:abcabcabcabc…… #include <cstdio> #include <cstring> using namespace std; const char str[] = "aababb"; int m…

题意:问是否能把MI通过以下规则转换成给定的字符串s. 1.使M之后的任何字符串加倍(即,将Mx更改为Mxx). 例如:MIU到MIUIU.2.用U替换任何III.例如:MUIIIU至MUUU.3.去掉任何UU. 例如:MUUU到MU. 分析: 1.MI的变换首先要复制I,可以复制为1,2,4,8,16,32,……(2的n次方)个. 2.由于可以用U替换任何III,所以将字符串s中所有的U变为I后,统计I的个数cnt. 3.由于可以去掉任何UU,所以转换成功必须满足cnt+6x==2的n次方.…

If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number. You are required to count the number of good numbers in the range from A to B, inclusive. InputThe first line has a number T (T <=…

http://acm.hdu.edu.cn/showproblem.php?pid=2855 化简这个公式,多写出几组就会发现规律 d[n]=F[2*n] 后面的任务就是矩阵快速幂拍一个斐波那契模板出来了 这里用的是2维 vector #include<iostream> #include<cstdio> #include<vector> using namespace std; typedef vector<int>vec; typedef vector&…