nefuoj Garden visiting 628 （组合数学&&Lucas定理）

Garden visiting

description

There is a very big garden at Raven’s residence. We regard the garden as an n*m rectangle. Raven’s house is at the top left corner, and the exit of the garden is at the bottom right. He can choose to take one step to only one direction (up, down, left or right) each time. Raven wants to go out of the garden as quickly as possible, so he wonders how many routes he could choose.
Raven knows there are many possible routes, so he only wants to know the number, which is the result that the total number of possible routes modes a given value p. He knows it is a simple question, so he hopes you may help him to solve it.

input

The first line of the input contains an integer T, which indicates the number of test cases.
Then it is followed by three positive integers n, m and p (1 <= n, m, p <= 10^5), showing the length and width of the garden and p to be the mod of the result.

output

For each case, output one number to show the result (the sum modes p).

sample_input

3
2 2 5
2 6 16
6 6 24

sample_output

2
6
12

hint

Sample 1: There are 2 routes in total.
Sample 2: There are 6 routes in total.

Sample 3: There are 252 routes in total.

#include<stdio.h>#include<string.h>#include<algorithm>#define ll long long#define N 200010using namespace std;int n,m,p;int P;bool pp;int pnum;void getp(){int i,j;for(i=2;i<N;i++){if(!pp[i]){P[pnum++]=i;for(j=i+i;j<N;j+=i)pp[j]=true;}}}int getzs(int n,int k){int ans=0;while(n>=k){n/=k;ans+=n;}return ans;}ll ksm(ll x,ll y)//快速幂{ll ans=1;while(y){if(y&1)ans=(ans*x)%p;x*=x;x%=p;y>>=1;}return ans;}int main(){getp();int t,i;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&p);n+=m-2;m--;ll ans=1;for(i=0;i<pnum&&P[i]<=n;i++){int num=getzs(n,P[i])-getzs(m,P[i])-getzs(n-m,P[i]);ans=ans*ksm(P[i],num)%p;}printf("%d\n",(int)ans);}return 0;}