zoukankan      html  css  js  c++  java
  • hdu 3037——Saving Beans

    Saving Beans

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 8258    Accepted Submission(s): 3302


    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.
    这题涉及乘法逆元、Lucas定理、组合数公式。
    通过分析题目可以得到方案数为C(n+m-1,n-1)%p
    再通过化简得:方案数为C(m+n,m)%p
     
    由于考虑到T的值很大,p的值较小,我们要对1...p 的阶乘打表
    另外,一定要注意0! = 1  !!!
    血的教训
     
    代码:
     1 #include <cstdio>
     2 #include <iostream>
     3 using namespace std;
     4 long long n,m,p;
     5 long long sum[500050];
     6 
     7 long long quick_mod(long long a,long long b)
     8 {
     9     long long ans=1;
    10     while(b)
    11     {
    12         if(b&1)ans=ans*a%p;
    13         b>>=1;
    14         a=a*a%p;
    15     }
    16     return ans;
    17 }
    18 
    19 long long C(long long n,long long m)
    20 {
    21     if(n<m)return 0;
    22     return (sum[n]*quick_mod((sum[m]*sum[n-m])%p,p-2)%p)%p;
    23 }
    24 
    25 long long Lucas(long long n,long long m)
    26 {
    27     if(m==0)return 1;
    28     return C(n%p,m%p)*Lucas(n/p,m/p)%p;
    29 }
    30 
    31 int main()
    32 {
    33     int cas=0;scanf("%d",&cas);
    34     while(cas--)
    35     {
    36         scanf("%I64d%I64d%I64d",&n,&m,&p);
    37         sum[0]=1;for(int i=1;i<=p;i++)sum[i]=sum[i-1]*i%p;
    38         printf("%I64d
    ",Lucas(n+m,m));
    39     }
    40     return 0;
    41 }
    View Code

     点个赞吧↓

  • 相关阅读:
    dedecms如何调用当前栏目的子栏目及子栏目文章
    dedecms调用当前栏目的子栏目怎么操作
    dedecms如何增加自定义字段
    关于朋友圈你所不知道的内幕
    dedecms如何快速删除跳转的文章(记得清空内容回收站)
    帝国cms调用栏目自定义字段(栏目简介)如何操作
    Introduction To Monte Carlo Methods
    Solr学习笔记-在Tomcat上部署执行Solr
    POJ 2029--Get Many Persimmon Trees +DP
    SNMP协议总结
  • 原文地址:https://www.cnblogs.com/lzxzy-blog/p/10320034.html
Copyright © 2011-2022 走看看