快速幂取模 POJ 3761 bubble sort

 

题目传送门

 1 /*
 2     题意：求冒泡排序扫描k次能排好序的全排列个数
 3     数学：这里有一个反序列表的概念，bj表示在j左边，但大于j的个数。不多说了，我也是看网上的解题报告。
 4     详细解释：http://blog.youkuaiyun.com/cscj2010/article/details/7820906
 5 */
 6 #include <cstdio>
 7 #include <algorithm>
 8 #include <cstring>
 9 #include <cmath>
10 using namespace std;
11 
12 typedef long long ll;
13 const int MAXN = 1e6 + 10;
14 const int INF = 0x3f3f3f3f;
15 const int MOD = 20100713;
16 ll fact[MAXN];
17 
18 void solve(void)    {
19     fact[0] = fact[1] = 1;
20     for (int i=2; i<=1000000; ++i)  {
21         fact[i] = fact[i-1] * i % MOD;
22     }
23 }
24 
25 ll pow_mod(ll x, ll n)  {
26     ll ret = 1;
27     while (n)   {
28         if (n & 1)  ret = ret * x % MOD;
29         x = x * x % MOD;
30         n >>= 1;
31     }
32     return ret;
33 }
34 
35 int main(void)  {       //POJ 3761 bubble sort
36     solve ();
37     int T;  scanf ("%d", &T);
38     while (T--) {
39         ll n, k;    scanf ("%I64d%I64d", &n, &k);
40         printf ("%I64d\n", (pow_mod (k + 1, n - k) - pow_mod (k, n - k) + MOD) % MOD * fact[k] % MOD);
41     }
42 
43     return 0;
44 }

 

转载于:https://www.cnblogs.com/Running-Time/p/4693102.html