HDU 4869 Turn the pokers

题目链接

题意：给定n个翻转扑克方式，每次方式对应可以选择其中xi张进行翻转，一共有m张牌，问最后翻转之后的情况数

思路：对于每一些翻转，如果能确定最终正面向上张数的情况，那么所有的情况就是所有情况的C(m, 张数)之和，那么这个张数进行推理会发现，其实会有一个上下界，每隔2个位置的数字就是可以的方案，因为在翻牌的时候，对应的肯定会有牌被翻转，而如果向上牌少翻一张，向下牌就要多翻一张，奇偶性是不变的，因此只要每次输入张数，维护上下界，最后在去求和即可

代码：

#include 
  
   
#include 
   
     typedef long long ll; const ll MOD = 1000000009; const int N = 100005; int n, m, num; ll fac[N]; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) {x = 1; y = 0; return a;} ll d = exgcd(b, a % b, y, x); y -= a / b * x; return d; } ll inv(ll a, ll n) { ll x, y; exgcd(a, n, x, y); return (x + n) % n; } ll C(int n, int m) { return fac[n] * inv(fac[m] * fac[n - m] % MOD, MOD) % MOD; } int main() { fac[0] = 1; for (ll i = 1; i < N; i++) fac[i] = fac[i - 1] * i % MOD; while (~scanf("%d%d", &n, &m)) { scanf("%d", &num); int up = num; int down = num; for (int i = 1; i < n; i++) { scanf("%d", &num); int up2 = m - down; int down2 = m - up; if (num >= down && num <= up) down = ((down&1)^(num&1)); else if (num < down) down = down - num; else down = num - up; if (num >= down2 && num <= up2) { up = m - ((up2&1)^(num&1)); } else if (num < down2) { up = m - (down2 - num); } else up = m - (num - up2); } ll ans = 0; for (int i = down; i <= up; i += 2) { ans = (ans + C(m, i)) % MOD; } printf("%lld\n", ans); } return 0; }