BZOJ1026: [SCOI2009]windy数

传送门

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;
}