poj—— 3037 Saving Beans

　　　　　　　　　　　　Saving Beans

　　　　　　　　　　　　Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
　　　　　　　　　　　　　　　　　　Total Submission(s): 5769    Accepted Submission(s): 2316

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

2009 Multi-University Training Contest 13 - Host by HIT

 

 

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题意：在n树中有多少种方式可以节省不超过m豆子（他们是一样的）。

思路：

题目可以转换成  x1+x2+……+xn=m 有多少组解，m在题中可以取0~m。

利用插板法可以得出x1+x2+……+xn=m解的个数为C（n+m-1，m）；

则题目解的个数可以转换成求   sum=C（n+m-1，0）+C（n+m-1，1）+C（n+m-1，2）……+C（n+m-1，m）

利用公式C（n，r）=C（n-1，r）+C（n-1，r-1）  == >  sum=C（n+m，m）。

就是要求C（n+m，m）%p。

代码：

#include
#include
#include
#include
#include
#define ll long long
using namespace std;
ll t,n,m,p,ans;
ll read()
{
    ll x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
ll qpow(ll n,ll k)
{
    ll res=1;
    while(k)
    {
        if(k&1) res=res*n%p;
        n=n*n%p; k>>=1;
    }return res;
}
ll c(ll n,ll m)
{
    if(m>n) return 0;
    ll n1=1,m1=1;
    for(ll i=n-m+1;i<=n;i++)
     n1=n1*i%p;
    for(ll i=1;i<=m;i++)
     m1=m1*i%p;
    return n1*qpow(m1,p-2);
}
ll lus(ll n,ll m)
{
     if(m==0) return 1;
     return c(n%p,m%p)*lus(n/p,m/p)%p;  
}
int main()
{
    t=read();
    while(t--)
    {
        n=read(),m=read(),p=read();
        ans=lus(n+m,m);
        printf("%lld\n",ans);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/z360/p/7360435.html