Saving Beans
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8258 Accepted Submission(s): 3302
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.
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.
1 2 5
2 1 5
3
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.
#include <cstdio>
#include <iostream>
using namespace std;
long long n,m,p;
long long sum[]; long long quick_mod(long long a,long long b)
{
long long ans=;
while(b)
{
if(b&)ans=ans*a%p;
b>>=;
a=a*a%p;
}
return ans;
} long long C(long long n,long long m)
{
if(n<m)return ;
return (sum[n]*quick_mod((sum[m]*sum[n-m])%p,p-)%p)%p;
} long long Lucas(long long n,long long m)
{
if(m==)return ;
return C(n%p,m%p)*Lucas(n/p,m/p)%p;
} int main()
{
int cas=;scanf("%d",&cas);
while(cas--)
{
scanf("%I64d%I64d%I64d",&n,&m,&p);
sum[]=;for(int i=;i<=p;i++)sum[i]=sum[i-]*i%p;
printf("%I64d\n",Lucas(n+m,m));
}
return ;
}
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