【codechef】Zombinatorial（找规律）
You are given a function f which is defined as :
Your task is to find the value of
where M is given in input.
Input Format
First line contains T, the number of test cases.
First line of each test case contain 3 space separated integers N, M and Q.
Next Q line follows, each line contain r.
Output Format
For each test case, output Q lines, each line containing the required answer.
Constraints
2 ≤ N ≤ 106
1 ≤ M ≤ 109
2 ≤ Sum of N over all test cases ≤ 106
1 ≤ Sum of Q over all test cases ≤ 2*105
1 ≤ T ≤ 105
1 < r < N
Sample Input 2 5 114 1 2 50 3874 3 31 17 21 Sample Output 72 3718 624 1144
Explanation for first test case
f[1] = 1
f[2] = 2
f[3] = 1*22 * 3 = 12
f[4] =1*23*32*4 = 8*9*4 = 288
f[5] = 1*24*33*42*5 =34560
value of f[5] / (f[2]*f[3]) = 1440 and 1440 %114 is 72
http://www.codechef.com/problems/FOMBRO
分奇偶两种情况考虑。
奇数(9):
19 28 37 46 55 64 73 82 91
19 21 31 41 51 61 71 81 91
19 21 32 42 52 62 72 82 91
19 21 32 43 53 63 73 82 91
19 21 32 43 54 64 73 82 91
偶数(8):
18 27 36 45 54 63 72 81
18 21 31 41 51 61 71 81
18 21 32 42 52 62 72 81
18 21 32 43 53 63 72 81
18 21 32 43 54 63 72 81
由此，规律应该很好看懂了吧~
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<map>
#include<cmath>
#include<vector>
#include<stdlib.h>
#include<cstdio>
#define ll long long
using namespace std;
ll add[1000001];
ll x[1000001];
int main(){
int t;
cin>>t;
while(t--){
ll n,m,q,a;
scanf("%lld %lld %lld",&n,&m,&q);
ll p=n/2+1,l,r;
if(n%2==0){
l=p;r=p;
add[0]=p%m;
}
else{
l=p;r=p+1;
add[0]=((p%m)*((p+1)%m))%m;
}
int k=0;
for(int i=l-1,j=r+1;j<=n;--i,++j){
add[++k]=(((add[k-1]%m)*(i%m))%m*(j%m))%m; //这里取模WA了好多下。。。以后注意
}
x[0]=1%m;
for(int i=1;i<=n/2;++i){
x[i]=((x[i-1]%m)*(add[k]%m))%m;
k--;
}
while(q--){
scanf("%lld",&a);
if(a>n/2)
a=n-a;
printf("%lld\n",x[a]);
}
}
return 0;
}