501D Misha and Permutations Summation 数据结构+打脸题

标签：

都快退役啦，小白书上的例题还不会。

给出一个序列S，则S的字典序为 sigma(dp[i] * (n-i)!) (1 <= i <= n)。dp[i] 表示[i，n]这一段S的子序列内比num[i]小的数字的个数。

对于两个序列A，B，(ord(A) + ord(B))%n!可以转化成 sigma( (A_dp[i] + B_dp[i] + (A_dp[i+1]+B_dp[i+1])/(n-i+1))%(n-i+1)  )( 1 <= i <= n ，可以认为dp[n+1] == 0)。

解码部分和编码部分为逆操作，详见代码。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>
#include <iomanip>

#pragma comment(linker,"/STACK:1024000000");
#define EPS (1e-6)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define INF 0x3f3f3f3f
#define Mod 1000000007
#define mod 1000000007

/** I/O Accelerator Interface .. **/
#define g (c=getchar())
#define d isdigit(g)
#define p x=x*10+c-'0'
#define n x=x*10+'0'-c
#define pp l/=10,p
#define nn l/=10,n
template<class T> inline T& RD(T &x)
{
    char c;
    while(!d);
    x=c-'0';
    while(d)p;
    return x;
}
template<class T> inline T& RDD(T &x)
{
    char c;
    while(g,c!='-'&&!isdigit(c));
    if (c=='-')
    {
        x='0'-g;
        while(d)n;
    }
    else
    {
        x=c-'0';
        while(d)p;
    }
    return x;
}
inline double& RF(double &x)      //scanf("%lf", &x);
{
    char c;
    while(g,c!='-'&&c!='.'&&!isdigit(c));
    if(c=='-')if(g=='.')
        {
            x=0;
            double l=1;
            while(d)nn;
            x*=l;
        }
        else
        {
            x='0'-c;
            while(d)n;
            if(c=='.')
            {
                double l=1;
                while(d)nn;
                x*=l;
            }
        }
    else if(c=='.')
    {
        x=0;
        double l=1;
        while(d)pp;
        x*=l;
    }
    else
    {
        x=c-'0';
        while(d)p;
        if(c=='.')
        {
            double l=1;
            while(d)pp;
            x*=l;
        }
    }
    return x;
}
#undef nn
#undef pp
#undef n
#undef p
#undef d
#undef g
using namespace std;

int st[800100];

int num[200100];

int ans[200100];

int Update(int site,int l,int r,int x,int val)
{
    if(l == r)
        return st[site] = val;
    int mid = (l+r)>>1;

    if(x <= mid)
        Update(site<<1,l,mid,x,val);
    else
        Update(site<<1|1,mid+1,r,x,val);

    return st[site] = st[site<<1] + st[site<<1|1];
}

int Query(int site,int L,int R,int l,int r)
{
    if(L == l && R == r)
        return st[site];

    int mid = (L+R)>>1;

    if(r <= mid)
        return Query(site<<1,L,mid,l,r);
    if(mid < l)
        return Query(site<<1|1,mid+1,R,l,r);

    return Query(site<<1,L,mid,l,mid) + Query(site<<1|1,mid+1,R,mid+1,r);
}

int BS(int s,int e,int x)
{
    int ans,mid,site,n = e;

    while(s <= e)
    {
        mid = (s+e)>>1;

        ans = Query(1,1,n,1,mid);

        if(ans <= x)
            s = mid+1;
        else
            site = mid,e = mid-1;
    }
    return site;
}

int main()
{
    int n,i,j,site;

    while(scanf("%d",&n) != EOF)
    {

        memset(ans,0,sizeof(ans));

        for(i = 1;i <= n; ++i)
            scanf("%d",&num[i]),++num[i];

        memset(st,0,sizeof(st));

        for(i = n;i >= 1; --i)
        {
            Update(1,1,n,num[i],1);
            ans[i] += Query(1,1,n,1,num[i])-1;
        }

        for(i = 1;i <= n; ++i)
            scanf("%d",&num[i]),++num[i];

        memset(st,0,sizeof(st));

        for(i = n;i >= 1; --i)
        {
            Update(1,1,n,num[i],1);
            ans[i] += Query(1,1,n,1,num[i])-1;
        }

        for(i = n;i >= 1; --i)
            if(ans[i] >= n-i+1)
                ans[i-1] += 1,ans[i] %= (n-i+1);

        for(i = 1;i <= n; ++i)
        {
            printf("%d%c",(site = BS(1,n,ans[i]))-1,i == n ? '\n' : ' ');

            Update(1,1,n,site,0);
        }

    }

    return 0;
}

501D Misha and Permutations Summation 数据结构+打脸题

标签：

原文地址：http://blog.csdn.net/zmx354/article/details/42743541