BZOJ2982: combination Lucas模板
2982: combination
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 734 Solved: 437
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Description
LMZ有n个不同的基友,他每天晚上要选m个进行[河蟹],而且要求每天晚上的选择都不一样。那么LMZ能够持续多少个这样的夜晚呢?当然,LMZ的一年有10007天,所以他想知道答案mod 10007的值。(1<=m<=n<=200,000,000)
Input
第一行一个整数t,表示有t组数据。(t<=200)
接下来t行每行两个整数n, m,如题意。
Output
T行,每行一个数,为C(n, m) mod 10007的答案。
Sample Input
4
5 1
5 2
7 3
4 2
5 1
5 2
7 3
4 2
Sample Output
5
10
35
6
10
35
6
HINT
Source
lucas模板题
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <queue> 7 #include <vector> 8 #include <map> 9 #include <string> 10 #include <cmath> 11 #include <sstream> 12 #define min(a, b) ((a) < (b) ? (a) : (b)) 13 #define max(a, b) ((a) > (b) ? (a) : (b)) 14 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a)) 15 template<class T> 16 inline void swap(T &a, T &b) 17 { 18 T tmp = a;a = b;b = tmp; 19 } 20 inline void read(long long &x) 21 { 22 x = 0;char ch = getchar(), c = ch; 23 while(ch < '0' || ch > '9') c = ch, ch = getchar(); 24 while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar(); 25 if(c == '-') x = -x; 26 } 27 const long long INF = 0x3f3f3f3f; 28 const long long MAXN = 200000000; 29 const long long MOD = 10007; 30 31 long long f[MOD + 1000], t, n, m; 32 long long pow(long long a, long long b) 33 { 34 long long r = 1, base = a % MOD; 35 for(;b;b >>= 1) 36 { 37 if(b & 1) r *= base, r %= MOD; 38 base *= base, base %= MOD; 39 } 40 return r; 41 } 42 long long ni(long long x) 43 { 44 return pow(x, MOD - 2); 45 } 46 long long C(long long n, long long m) 47 { 48 if(n < m) return 0; 49 int tmp = f[n] * ni(f[m]) % MOD , tmp2 = tmp * ni(f[n - m]) % MOD; 50 return tmp2; 51 } 52 long long lucas(long long n, long long m) 53 { 54 if(!m) return 1; 55 else return C(n % MOD, m % MOD) * lucas(n / MOD, m / MOD) % MOD; 56 } 57 58 int main() 59 { 60 read(t);f[0] = 1; 61 for(long long i = 1;i <= MOD;++ i) 62 f[i] = f[i - 1] * i % MOD; 63 for(long long i = 1;i <= t;++ i) 64 { 65 read(n), read(m); 66 printf("%lld\n", lucas(n, m)); 67 } 68 return 0; 69 }