【BZOJ2729】排队（组合数学 + 高精度）

传送门

    HNOI2012 排队
    题意：某中学有 n 名男同学，m 名女同学和两名老师要排队参加体检。他们排成一条直线，并且任意两名女同学不能相邻，两名老师也不能相邻，那么一共有多少种排法呢?（任意两人不同）

I think

    排列组合+高精度
    考虑先放男生,后放老师,再放女生

    先用男生把两个老师隔开,再用男生和老师把所有女生隔开
    方案数为: A(n,n)*A(n+1,2)*A(n+3,m)

    或者用女生隔开老师
    方案数为:A(n,n)*A(m,1)*A(2,2)*A(n+1,1)*A(n+3,m-1)(此时两个老师视为整体)

    注意排列公式: A(m,n) = m!/(m-n)!

Code

#include<cstdio>#include<cstring>using namespace std;typedef long long LL;const int sm = 2e3 + 10,Mod = 1e8;int n,m;int Max(int x,int y) { return x>y?x:y; }struct Bign {    int len;    LL val[sm];    Bign() { len = 1; memset(val,0,sizeof(val)); val[0] = 1; }    void init() { val[0] = 1;}    Bign operator * (const int x) {        Bign ret; LL t = 0;        ret.val[0] = 0, ret.len = len + 1;         for(int i = 0; i < ret.len; ++i) {            ret.val[i] = (val[i] * x + t) % Mod;            t = (val[i] * x + t) / Mod;        }        while(ret.val[ret.len-1] == 0 && ret.len > 1) --ret.len;        return ret;    }    Bign operator + (const Bign & x) {        Bign ret;  ret.len = Max(len,x.len) + 1;        LL t = 0; ret.val[0] = 0;        for(int i = 0; i < ret.len; ++i) {            ret.val[i] = (val[i] + x.val[i] + t) % Mod;            t = (val[i] + x.val[i] + t) / Mod;        }        while(ret.val[ret.len-1]==0 && ret.len>1)             --ret.len;        return ret;    }    void print() {        printf("%lld",val[len-1]);        for(int i = len-2; i >= 0; --i)            printf("%08lld",val[i]);        putchar(10);    }}a,b;int main() {    bool flaga = 0, flagb = 0;    scanf("%d%d",&n,&m);    if(n+3<m||n+m+1<2) return puts("0"),0;    else {        if(2<=n+1 && m<=n+3) {            flaga = 1;            for(int i = 1; i <= n+3; ++i) {                if(i <= n) a = a * i;                if(i > n-m+3) a = a * i;            }            a = a * (n*(n+1));        }        if(m!=0 && m-1<=n+2) {            flagb = 1;            for(int i = 1; i <= n+2; ++i) {                if(i <= n) b = b * i;                if(i > n-m+3) b = b * i;            }            b = b * (2*m*(n+1));        }        if(!flaga) a.val[0] = 0;        if(flagb) a = a + b;        a.print();     }    return 0;}