題目鏈接

BZOJ2729

題解

高考數(shù)學題。。。

我們先把老師看做男生，女生插空站
如果兩個老師相鄰，我們把他們看做一個男生，女生插空站
對于\(n\)個男生\(m\)個女生的方案數(shù)：
\[n!m!{n + 1 \choose m}\]

還要特判一下沒有男生女生的情況

然后高精即可

#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<map> #define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt) #define REP(i,n) for (int i = 1; i <= (n); i++) #define mp(a,b) make_pair<int,int>(a,b) #define cls(s) memset(s,0,sizeof(s)) #define cp pair<int,int> #define LL long long int using namespace std; const int maxn = 100005,B = 10000,maxm = 100005,INF = 1000000000; inline int read(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag; } struct NUM{int s[4000],len;NUM(){cls(s); len = 0;}void out(){printf("%d",s[len]);for (int i = len - 1; i; i--)printf("%04d",s[i]);} }A,C; inline NUM operator *(const NUM& a,const int& b){NUM c;c.len = a.len;LL carry = 0,tmp;for (int i = 1; i <= a.len; i++){tmp = a.s[i] * b + carry;c.s[i] = tmp % B;carry = tmp / B;}while (carry) c.s[++c.len] = carry % B,carry /= B;return c; } inline NUM operator -(const NUM& a,const NUM& b){NUM c;c.len = a.len;int carry = 0,tmp;for (int i = 1; i <= a.len; i++){tmp = a.s[i] - b.s[i] + carry;if (tmp < 0) tmp += B,carry = -1;else carry = 0;c.s[i] = tmp;}while (c.len && !c.s[c.len]) c.len--;return c; } int main(){int n = read(),m = read();if (n + 3 < m || (!n && !m)){puts("0"); return 0;}n += 2;A.s[A.len = 1] = 1;for (int i = n; i > n - m + 1; i--)A = A * i;for (int i = n + 1; i; i--)A = A * i;n--;if (n + 1 >= m){C.s[C.len = 1] = 1;for (int i = n; i > n - m + 1; i--)C = C * i;for (int i = n + 1; i; i--)C = C * i;C = C * 2;}A = A - C;A.out();return 0; }

轉載于:https://www.cnblogs.com/Mychael/p/9072096.html

總結

以上是生活随笔為你收集整理的BZOJ2729 [HNOI2012]排队 【高精 + 组合数学】的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。