hdu 4869
Turn the pokers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 676 Accepted Submission(s): 231
Problem Description
During summer vacation,Alice stay at home for a long time, with nothing to do. She went out and bought m pokers, tending to play poker. But she hated the traditional gameplay. She wants to change. She puts these pokers face down, she decided to flip poker n times, and each time she can flip Xi pokers. She wanted to know how many the results does she get. Can you help her solve this problem?
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1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 6 using namespace std; 7 8 typedef long long ll; 9 10 const int MAX = 1e5 + 7; 11 const int MOD = 1e9 + 9; 12 int N, M; 13 int x[MAX]; 14 ll c[MAX]; 15 16 ll mod_pow(ll x, ll n, ll mod) { 17 ll res = 1; 18 while (n > 0) { 19 if (n & 1) res = res * x % mod; 20 x = x * x % mod; 21 n >>= 1; 22 } 23 24 return res; 25 } 26 27 void solve() { 28 int Min = 0, Max = 0; 29 for (int i = 1; i <= N; ++i) { 30 int t1, t2; 31 if (Min >= x[i]) { 32 t1 = Min - x[i]; 33 } else if (x[i] <= Max) { 34 t1 = !((Min & 1) == (x[i] & 1)); 35 } else { 36 t1 = x[i] - Max; 37 } 38 39 if (Max + x[i] <= M) { 40 t2 = Max + x[i]; 41 } else if (Min + x[i] <= M) { 42 t2 = (((Min + x[i]) & 1) == (M & 1)) ? M : M - 1; 43 } else { 44 t2 = 2 * M - (Min + x[i]); 45 } 46 47 Min = t1; Max = t2; 48 } 49 50 //printf("min = %d max = %d\n", Min, Max); 51 ll ans = 0; 52 c[0] = 1; 53 if (Min == 0) ans += 1; 54 for (int i = 1; i <= Max; ++i) { 55 if (M - i < i) c[i] = c[M - i]; 56 else 57 c[i] = c[i - 1] * (M - i + 1) % MOD * mod_pow(i, MOD - 2,MOD) % MOD; 58 if (i >= Min && (i & 1) == (Min & 1)) ans = (ans + c[i]) % MOD; 59 60 } 61 62 printf("%I64d\n", ans); 63 } 64 int main() 65 { 66 while (scanf("%d%d", &N, &M) != EOF) { 67 for (int i = 1; i <= N; ++i) { 68 scanf("%d", &x[i]); 69 } 70 71 solve(); 72 } 73 //cout << "Hello world!" << endl; 74 return 0; 75 }
Input
The input consists of multiple test cases.
Each test case begins with a line containing two non-negative integers n and m(0<n,m<=100000).
The next line contains n integers Xi(0<=Xi<=m).
Each test case begins with a line containing two non-negative integers n and m(0<n,m<=100000).
The next line contains n integers Xi(0<=Xi<=m).
Output
Output the required answer modulo 1000000009 for each test case, one per line.
Sample Input
3 4
3 2 3
3 3
3 2 3
Sample Output
8
3
Hint
For the second example:
0 express face down,1 express face up
Initial state 000
The first result:000->111->001->110
The second result:000->111->100->011
The third result:000->111->010->101
So, there are three kinds of results(110,011,101)
Author
FZU
Source
首先1个个数的奇偶性和翻牌次数的奇偶性相同,然后根据分类讨论找出出现的最少的1 和最多的1,最终的结果肯定是连续的奇数或者偶数
然后求出sum(c(M,i)) ( Min < i <= Max)连续的奇数或偶数
根据费马小定理可以知道若p为质数,gcd(a, p) = 1,则a ^ (p - 2) = 1 / p (mod p) 根据这个求出组合数就可以了
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