51nod 1799 二分答案(分块打表)

首先题目等价于求出满足运行二分程序后最后r=k的排列种数。

显然对于这样的二分程序，起作用的只有mid点，mid处的值决定了接下来要递归的子区间。

于是可以一遍二分求出有多少个mid点处的值<=m,有多少个mid点处的值>m,不妨设为x和y，

那么由组合数学可以推出最后的答案是 C(x,m)*C(y,n-m)*(n-x-y)!%MOD.

由于x和y很小，所以前面两个组合数可以暴力算出来。而后面的阶乘显然是不能直接求的。

打表的话n<=1e9，显然会MLE，于是把n分成100块，预处理出n=1e7，2e7...3e7...1e9的答案。

然后在块内暴力即可。

显然时间复杂度为O(logn+n/100).

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-8
# define MOD 1000000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=105;
//Code begin...

int mod[N]={
    1,924724006,582347126,500419162,881147799,693776109,435873621,279027658,727951124,398578768,678364145,204828554,345795998,116118093,359401113,236930793,856493327,207383191,617606889,933753281,26701748,329394893,360779992,416008308,187501984,165706817,328891607,16385287,117411011,404196042,765064133,239669664,761588352,566114869,673499119,840260100,352356536,53839501,178657924,373444237,227300165,207172723,444208499,367531373,297449176,605324209,729265513,567907756,125889461,250743107,666666670,598576559,632705086,295855233,185718228,414607857,737215408,863388390,182290465,707552496,881713600,417895708,490627919,364521407,775935292,972492338,473340273,920880265,530581,696910290,64037482,649527920,756691728,283805222,711255329,825205499,263679166,341083474,914727729,919247968,465317279,960145703,274813468,393588827,65909169,521964827,794328994,484551338,521297378,54488990,591837535,255746228,25827429,177799409,92011129,469664591,35708489,197025781,288851931,254032854
};
int get(int x){
    if (x==0) return 1;
    int i=mod[(x-1)/10000000], j=(x-1)%10000000;
    LL c=x/10000000*10000000+1;
    FOR(y,1,j) i=(LL)i*(c+y)%MOD;
    return i;
}
int main ()
{
    int n, m, k, x=0, y=0;
    scanf("%d%d%d",&n,&m,&k);
    int l=1, r=n, mid;
    while (l<=r) {
        mid=(l+r)>>1;
        if (mid<=k) l=mid+1, ++x;
        else r=mid-1, ++y;
    }
    LL ans=1;
    for (int i=m; i>=m-x+1; --i) ans=ans*i%MOD;
    for (int i=n-m; i>=n-m-y+1; --i) ans=ans*i%MOD;
    printf("%lld
",ans*get(n-x-y)%MOD);
    return 0;
}