[HDU] 3037 Saving Beans

Saving Beans
Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6279    Accepted Submission(s): 2537


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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题目可以转化为 n个盒子，m个球，允许空着不放，有多少种放法。
用神器的挡板法可以求得
当放了k(k<=m)个球，有C(k-n+1,n-1)种放法。
求和，利用杨辉三角递推式化简。
ANS=C(n-1,n-1)+C(n,n-1)+C(n+1,n-1)+……….+C(n+m-2,n-1)+C(n+m-1,n-1)
=C(n+m,n)

再用Lucas定理解决模p即可。
（原来的快速幂好像。。出现了问题？）

#include<iostream>
#include<cstdio>
using namespace std;

long long  t;

long long poww2(long long x,long long m,long long p){
    if(m==0) return 1;
    long long tmp=poww2(x,m/2,p);
    tmp=(tmp*tmp)%p;
    if(m%2==1) tmp=(tmp*x)%p;
    return tmp;
}


long long poww(long long a,long long b) {
    long long ans=1,base=a;
    if(b==0) return 1;
    while(b!=0) {
        if(b&1!=0)
            ans*=base;
        base*=base;
        b>>=1;
    }
    return ans;
}

long long c(long long m,long long n,long long p) {
    if(m<n) return 0;
    if(n>m-n) n=m-n;
    long long s1=1,s2=1;
    for(int i=1; i<=n; i++) {
        s1=s1*(m-i+1)%p;
        s2=s2*i%p;
    }
    return s1*poww2(s2,p-2,p)%p;
}

long long zc(long long x,long long y) {
    if(y==1) return x;
    return 1.0*x/y*zc(x-1,y-1);
}

long long lucas(long long m,long long n,long long p) {
    if(n==0) return 1;
    return c(m%p,n%p,p)*lucas(m/p,n/p,p)%p;
}

int main() {
    long long m,n,r;
    scanf("%lld",&t);
//  cin>>t;
    while(t--) {
    scanf("%lld%lld%lld",&m,&n,&r);
//  cin>>m>>n>>r;
//  cout<<"C: "<<zc(m,n);
//  cout<<endl<<"%: ";
//  cout<<m<<" "<<n<<endl;
    printf("%lld\n",lucas(m+n,m,r));
//  cout<<lucas(m+n,m,r)<<endl;
//  cout<<"STD:"<<zc(m+n,n)<<endl;;
    }
    return 0;
}

转载于:https://www.cnblogs.com/ghostcai/p/9247520.html