Saving Beans

Saving Beans

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5761    Accepted Submission(s):
2310

Problem Description
Although winter is far away, squirrels have to work day
and night to save beans. They need plenty of food to get through those long cold
days. After some time the squirrel family thinks that they have to solve a
problem. They suppose that they will save beans in n different trees. However,
since the food is not sufficient nowadays, they will get no more than m beans.
They want to know that how many ways there are to save no more than m beans
(they are the same) in n trees.

Now they turn to you for help, you should
give them the answer. The result may be extremely huge; you should output the
result modulo p, because squirrels can’t recognize large numbers.

 
Input
The first line contains one integer T, means the number
of cases.

Then followed T lines, each line contains three integers n, m,
p, means that squirrels will save no more than m same beans in n different
trees, 1 <= n, m <= 1000000000, 1 < p < 100000 and p is guaranteed
to be a prime.

 
Output
You should output the answer modulo p.
 
Sample Input
2
1 2 5
2 1 5
 
Sample Output
3
3

Hint

Hint

For sample 1, squirrels will put no more than 2 beans in one tree. Since trees are different, we can label them as 1, 2 … and so on.
The 3 ways are: put no beans, put 1 bean in tree 1 and put 2 beans in tree 1. For sample 2, the 3 ways are:
put no beans, put 1 bean in tree 1 and put 1 bean in tree 2.

 
Source
 
/*
题目相当于求n个数的和不超过m的方案数。
如果和恰好等于m,那么就等价于方程x1+x2+...+xn = m的解的个数,利用插板法可以得到方案数为:
(m+1)*(m+2)...(m+n-1) = C(m+n-1,n-1) = C(m+n-1,m)
现在就需要求不大于m的,相当于对i = 0,1...,m对C(n+i-1,i)求和,根据公式C(n,k) = C(n-1,k)+C(n-1,k-1)得
C(n-1,0)+C(n,1)+...+C(n+m-1,m)
= C(n,0)+C(n,1)+C(n+1,2)+...+C(n+m-1,m)
= C(n+m,m)
现在就是要求C(n+m,m) % p,其中p是素数。
然后利用Lucas定理的模板就可以轻松的求得C(n+m,m) % p的值
*/
#include<iostream>
#include<cstdio>
#include<cstring> #define N 100007 using namespace std;
long long f[N]; long long Mi(long long a,long long b,long long p)
{
long long res=;
while(b)
{
if(b&) res=res*a%p;
b>>=;a=a*a%p;
}return res;
} long long C(long long n,long long m,long long p)
{
if(m>n)return ;
return f[n]*Mi(f[m]*f[n-m]%p,p-,p)%p;
} long long Lcs(long long n,long long m,long long p)
{
if(m==)return ;
return (C(n%p,m%p,p)*Lcs(n/p,m/p,p))%p;
} int main()
{
long long n,m,p;long long t;
cin>>t;
while(t--)
{
cin>>n>>m>>p;
f[]=;
for(long long i=;i<=p;i++)
f[i]=f[i-]*i%p;
printf("%lld\n",Lcs(n+m,m,p));
}
return ;
}

hdu 3037Saving Beans(卢卡斯定理)的更多相关文章

  1. hdu3037Saving Beans——卢卡斯定理

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=3037 卢卡斯定理模板——大组合数的取模 代码如下: #include<iostream> #i ...

  2. 数学--数论--HDU 4675 GCD of Sequence(莫比乌斯反演+卢卡斯定理求组合数+乘法逆元+快速幂取模)

    先放知识点: 莫比乌斯反演 卢卡斯定理求组合数 乘法逆元 快速幂取模 GCD of Sequence Alice is playing a game with Bob. Alice shows N i ...

  3. hdu3037——卢卡斯定理

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=3037 卢卡斯定理模板——大组合数取模 #include<iostream> #include& ...

  4. 【BZOJ4403】序列统计(组合数学,卢卡斯定理)

    [BZOJ4403]序列统计(组合数学,卢卡斯定理) 题面 Description 给定三个正整数N.L和R,统计长度在1到N之间,元素大小都在L到R之间的单调不降序列的数量.输出答案对10^6+3取 ...

  5. 【Luogu3807】【模板】卢卡斯定理(数论)

    题目描述 给定\(n,m,p(1≤n,m,p≤10^5)\) 求 \(C_{n+m}^m mod p\) 保证\(P\)为\(prime\) \(C\)表示组合数. 一个测试点内包含多组数据. 输入输 ...

  6. 【数论】卢卡斯定理模板 洛谷P3807

    [数论]卢卡斯定理模板 洛谷P3807 >>>>题目 [题目] https://www.luogu.org/problemnew/show/P3807 [输入格式] 第一行一个 ...

  7. 【XSY2691】中关村 卢卡斯定理 数位DP

    题目描述 在一个\(k\)维空间中,每个整点被黑白染色.对于一个坐标为\((x_1,x_2,\ldots,x_k)\)的点,他的颜色我们通过如下方式计算: 如果存在一维坐标是\(0\),则颜色是黑色. ...

  8. 【CTSC2017】【BZOJ4903】吉夫特 卢卡斯定理 DP

    题目描述 给你一个长度为\(n\)的数列\(a\),求有多少个长度\(\geq 2\)的不上升子序列\(a_{b_1},a_{b_2},\ldots,a_{b_k}\)满足 \[ \prod_{i=2 ...

  9. 卢卡斯定理&扩展卢卡斯定理

    卢卡斯定理 求\(C_m^n~mod~p\) 设\(m={a_0}^{p_0}+{a_1}^{p_1}+\cdots+{a_k}^{p_k},n={b_0}^{p_0}+{b_1}^{p_1}+\cd ...

随机推荐

  1. Html test

    <!DOCTYPE html> <html lang="en" xmlns="http://www.w3.org/1999/xhtml"> ...

  2. Python 之糗事百科多线程爬虫案例

    import requests from lxml import etree import json import threading import queue # 采集html类 class Get ...

  3. 怎么选择最适合自己的Python培训机构?

    Python培训已经成为入门Python的一个重要途径,它的优势在于学习知识的系统性.快速性和实用性.Python培训毕业的学员大多数拥有较强的实战动手能力,能够较快上手,更符合企业需求. 不过,大部 ...

  4. 在vue中通过js动态控制图片按比列缩放

    1.html 通过外层的div来给img对应的class,隐藏的img是得到img图片请求回来时的原始尺寸.外层div是固定大小,因此,图片有两种情况去适应外部div的尺寸.一种是宽度大于高度的情况, ...

  5. recreate dbcontrol on database 11.2.0.1 using emca

    [oracle@osb ~]$ env | grep ORA ORACLE_SID=ACE ORACLE_BASE=/oracle ORACLE_TERM=xterm ORACLE_HOME=/ora ...

  6. Getmemory问题

    题目一: [cpp] view plaincopy void GetMemory( char *p ) { p = ( ); } void Test( void ) { char *str = NUL ...

  7. 准备MPI编程环境——Visual Studio

    准备 下载并安装Visual Studio 2017 下载并安装MPI  (建议使用MSMPI,相对简单方便一点,可以从微软官网下载获得) 配置 新建空白项目 在该项目中新建源文件 右击项目-> ...

  8. 【codeforces 509C】Sums of Digits

    [题目链接]:http://codeforces.com/contest/509/problem/C [题意] 给你一个数组b[i] 要求一个严格升序的数组a[i]; 使得a[i]是b[i]各个位上的 ...

  9. Wizard's Tour

    F. Wizard's Tour time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...

  10. Eleastisearch6.0.0 read_only_allow_delete: false

    Eleastisearch6.0.0由单节点升级到多节点集群cluster时候出现的分片同步错误问题解决 原创 2018年01月18日 16:33:21 5 启动多个节点的ES后,ES开始推举mast ...