Saving Beans
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2523 Accepted Submission(s): 933
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.
2 1 2 5 2 1 5
3 3HintHint 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.
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
下面简单介绍一下Lucas定理:
Lucas定理是用来求 C(n,m) mod p的值,p是素数(从n取m组合,模上p)。
描述为:
Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p)
Lucas(x,0,p)=1;
简单的理解就是:
以求解n! % p 为例,把n分段,每p个一段,每一段求得结果是一样的。但是需要单独处理每一段的末尾p,2p,...,把p提取出来,会发现剩下的数正好又是(n/p)! ,相当于划归了一个子问题,这样递归求解即可。
这个是单独处理n!的情况,当然C(n,m)就是n!/(m! *(n-m)!),每一个阶乘都用上面的方法处理的话,就是Lucas定理了
Lucas最大的数据处理能力是p在10^5左右。
而C(a,b) =a! / ( b! * (a-b)! ) mod p
其实就是求 ( a! / (a-b)!) * ( b! )^(p-2) mod p
(上面这一步变换是根据费马小定理:假如p是质数,且a,p互质,那么a的(p-1)次方除以p的余数恒为1,
那么a和a^(p-2)互为乘法逆元,则(b / a) = (b * a^(p-2) ) mod p)
#include<stdio.h>
#define N 100005
typedef __int64 ll;
ll f[N];
void init(ll p)
{
ll i;
f[0]=1;
for(i=1;i<=p;i++)
f[i]=f[i-1]*i%p;
}
ll powmod(ll a,ll b,ll p)
{
ll ans=1;
while(b)
{
if(b&1)
ans=ans*a%p;
a=a*a%p;
b/=2;
}
return ans;
}
ll lucas(ll n,ll m,ll p)
{
ll ans=1;
while(n&&m)
{
ll nn=n%p,mm=m%p;
if(nn<mm) return 0;
ans=ans*f[nn]*powmod(f[mm]*f[nn-mm]%p,p-2,p)%p;
n/=p;
m/=p;
}
return ans;
}
int main()
{
__int64 t,n,m,p;
scanf("%I64d",&t);
while(t--)
{
scanf("%I64d%I64d%I64d",&n,&m,&p);
init(p);
printf("%I64d\n",lucas(n+m,n,p));
}
return 0;
}