BZOJ1026: [SCOI2009]windy数

标签：枚举   else   dia   数位dp   break   scoi2009   init   its   ini   

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md直接wa了78次，身败名裂

没学过数位DP硬搞了一道数位DP的模板题，感觉非常的愉（sha）悦（cha）。

二分转化枚举思想。首先DP预处理出来$f[i][j]$表示有$i$位且第$i$位为$j$的windy数有多少个，然后搞个$g[i]$表示$i$位上可以有多少个windy数。然后二分出来最大的小于$A$和$B$的windy数。相减就好了。

//BZOJ 1026
//by Cydiater
//2016.10.24
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <map>
#include <iomanip>
#include <bitset>
using namespace std;
#define ll long long
#define up(i,j,n)       for(int i=j;i<=n;i++)
#define down(i,j,n)     for(int i=j;i>=n;i--)
const int oo=0x3f3f3f3f;
inline ll read(){
    char ch=getchar();ll x=0,f=1;
    while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
ll f[15][15],A,B,g[15],ten[15];
namespace solution{
    void init(){
        A=read();B=read();
        ll tmp=1;
        up(i,1,14){
            ten[i]=tmp;
            tmp*=10LL;
        }
    }
    ll col(ll pos,ll re,ll last){
        if(pos==0)return 0;
        if(last==-1){
			if(pos==1){
				if(re<=f[pos][0])return 0;
				re-=f[pos][0];
			}
            up(i,1,9){
                if(re<=f[pos][i])return  ten[pos]*i+col(pos-1,re,i);
                re-=f[pos][i];
            }
        }else{
			if(pos==1&&last-2>=0){
				if(re<=f[pos][0])return 0;
				re-=f[pos][0];
			}
            if(last-2>=0){
                ll sum=0;
                up(i,2,9)sum+=f[pos-1][i];
                if(re<=sum)      return col(pos-1,re,0);
                else            re-=sum;
            }
            up(i,1,9)if(abs(i-last)>=2){
                if(re<=f[pos][i])    return ten[pos]*i+col(pos-1,re,i);
                re-=f[pos][i];
            }
        }
    }
    ll check(ll id){
        ll last=-1,ans=0,high=0;
        down(i,11,1)if(g[i]<id){
            high=i;
            id-=g[i];
            break;
        }
        ans=col(high+1,id,-1);
        return ans;
    }
    ll get(ll num){
        ll leftt=0,rightt=10000000000LL,mid;
        while(leftt+1<rightt){
            mid=(leftt+rightt)>>1;
            if(check(mid)<=num)   leftt=mid;
            else                rightt=mid;
        }
        if(check(rightt)<=num)       return rightt;
        return leftt;
    }
    void slove(){
        memset(f,0,sizeof(f));
        up(i,0,9)f[1][i]=1;
        up(i,2,14)up(j,0,9){
            up(k,j+2,9)f[i][j]+=f[i-1][k];
            down(k,j-2,0)f[i][j]+=f[i-1][k];
        }
		if(A>B)swap(A,B);
        memset(g,0,sizeof(g));
        g[0]=1;
        up(i,1,14){
            g[i]+=g[i-1];
            up(j,1,9)g[i]+=f[i][j];
        }
        cout<<get(B)-get(A-1)<<endl;
    }
}
int main(){
	//freopen("input.in","r",stdin);
	//freopen("out1.out","w",stdout);
    using namespace solution;
    init();
    slove();
    return 0;
}

BZOJ1026: [SCOI2009]windy数

标签：枚举   else   dia   数位dp   break   scoi2009   init   its   ini   

原文地址：http://www.cnblogs.com/Cydiater/p/5994793.html