hannnnah_j’s Biological Test
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Problem Description
hannnnah_j is a teacher in WL High school who teaches biology.
One day, she wants to test m students, thus she arranges n different seats around a round table.
In order to prevent cheating, she thinks that there should be at least k empty seats between every two students.
hannnnah_j is poor at math, and she wants to know the sum of the solutions.So she turns to you for help.Can you help her? The answer maybe large, and you need to mod 1e9+7.
One day, she wants to test m students, thus she arranges n different seats around a round table.
In order to prevent cheating, she thinks that there should be at least k empty seats between every two students.
hannnnah_j is poor at math, and she wants to know the sum of the solutions.So she turns to you for help.Can you help her? The answer maybe large, and you need to mod 1e9+7.
Input
First line is an integer T(T≤1000).
The next T lines were given n, m, k, respectively.
0 < m < n < 1e6, 0 < k < 1000
The next T lines were given n, m, k, respectively.
0 < m < n < 1e6, 0 < k < 1000
Output
For each test case the output is only one integer number ans in a line.
Sample Input
2
4 2 6
5 2 1
Sample Output
0 5
思路：
1 #include <iostream> 2 #include <cstdio> 3 4 using namespace std; 5 typedef long long LL; 6 const int p = 1e9 + 7; 7 LL quick_mod(LL a, LL b) { 8 LL ans = 1; 9 a %= p; 10 while(b) { 11 if(b & 1) { 12 ans = ans * a % p; 13 b--; 14 } 15 b >>= 1; 16 a = a * a % p; 17 } 18 return ans; 19 } 20 21 LL C(LL n, LL m) { 22 if(m > n) return 0; 23 LL ans = 1; 24 for(int i=1; i<=m; i++) { 25 LL a = (n + i - m) % p; 26 LL b = i % p; 27 ans = ans * (a * quick_mod(b, p-2) % p) % p; 28 } 29 return ans; 30 } 31 LL Lucas(LL n, LL m) { 32 if(m == 0) return 1; 33 return C(n % p, m % p) * Lucas(n / p, m / p) % p; 34 } 35 int main() { 36 int k, T, n, m; 37 scanf("%d", &T); 38 while(T--){ 39 scanf("%d%d%d", &n, &m, &k); 40 if(m == 1) printf("%d ", n); 41 else { 42 printf("%d ", n*Lucas(n-k*m-1, m-1)%p*quick_mod(m, p-2)%p); 43 } 44 } 45 return 0; 46 }
看了看别人交的代码，发现自己的好慢