2016 ShenYang regional online 1007||hdu 5898 odd-even number (数位dp)

题目链接

题意:题意说得一点页不清楚。。。意思在询问在区间[l,r]中满足某条件的数。该条件是,该数的任何一段数字是奇数组成的数串必须有偶数长度,任何一段数字是偶数组成的数串必须由奇数长度。

对于样例1,满足条件的29个数字分别是: 2,4,6,8,11,13,15,17,19,31,33,35,37,39,51,53,55,57,59,71,73,75,77,79,91,93,95,97,99.

对于样例2,满足条件的36的数字分别是:

110,112,114,116,118,

130,132,134,136,138

150,152,154,156,158

170,172,174,176,178

190,192,194,196,198

200,202,204,206,208,220

211,213,215,217,219

思路:数位dp.dp[i][j][k]表示长度为i,奇偶性相同的连续由j个,上一个的奇偶性为k.

注意不允许前导0.

其他就是细节了,太久没写数位dp调了好久啊QAQ

  1/* ***********************************************
  2Author :111qqz
  3Created Time :Sun 18 Sep 2016 01:11:37 PM CST
  4File Name :code/net/2016/shenyang/1007.cpp
  5************************************************ */
  6#include <cstdio>
  7#include <cstring>
  8#include <iostream>
  9#include <algorithm>
 10#include <vector>
 11#include <queue>
 12#include <set>
 13#include <map>
 14#include <string>
 15#include <cmath>
 16#include <cstdlib>
 17#include <ctime>
 18#define fst first
 19#define sec second
 20#define lson l,m,rt<<1
 21#define rson m+1,r,rt<<1|1
 22#define ms(a,x) memset(a,x,sizeof(a))
 23typedef long long LL;
 24#define pi pair < int ,int >
 25#define MP make_pair
 26using namespace std;
 27const double eps = 1E-8;
 28const int dx4[4]={1,0,0,-1};
 29const int dy4[4]={0,-1,1,0};
 30const int inf = 0x3f3f3f3f;
 31int digit[40];
 32LL dp2[30][30][5];
 33LL dfs2(int pos,int cnt,int lst,bool limit,bool prehasnonzero)
 34{
 35    if (pos==0)
 36    {
 37	if (lst==1&&cnt%2==0) return 1;
 38       if (lst==2&&cnt%2==1) return 1;
 39	return 0;
 40    }
 41    if (prehasnonzero&&!limit&&dp2[pos][cnt][lst]!=-1) return dp2[pos][cnt][lst];
 42    int mx = limit?digit[pos]:9;
 43    LL res = 0 ;
 44    if (!prehasnonzero)
 45    {
 46	for ( int i = 0 ; i <= mx ; i++)
 47	{
 48	    res+=dfs2(pos-1,i==0?0:1,i==0?0:2-i%2,limit&&i==mx,i==0?false:true);
 49	}
 50    }
 51    else
 52    {
 53	for ( int i = 0 ; i <= mx ; i++)
 54	{
 55	   if (lst==1&&cnt>0&&cnt%2==0&&i%2==1&&pos==1) continue;
 56	    if (lst==1&&cnt%2==1&&i%2==0) continue;
 57	    if (lst==2&&cnt>0&&cnt%2==0&&i%2==1) continue;
 58	    if (lst==2&&cnt%2==1&&i%2==0&&pos==1) continue;
 59	    res+=dfs2(pos-1,((i%2==1&&lst==1)||(i%2==0&&lst==2))?cnt+1:1,2-i%2,limit&&i==mx,true);
 60	}
 61    }
 62    if (prehasnonzero&&!limit) dp2[pos][cnt][lst] = res;
 63    return res;
 64}
 65/*
 66LL dfs( int pos,int odd,int even,int lst,bool limit, bool prehasnonzero)
 67{
 68    if (pos==0) return 1;
 69    if (prehasnonzero&&!limit&&dp[pos][odd][even][lst]!=-1) return dp[pos][odd][even][lst];
 70    int mx = limit?digit[pos]:9;
 71    LL res = 0;
 72    if (prehasnonzero)
 73    {
 74	for ( int i = 0 ; i <= mx ; i++)
 75	{
 76	    if (lst==1)
 77	    {
 78		if (odd%2==0&&i%2==1) continue;
 79		if (odd%2==1&&i%2==0) continue;
 80	    }else if (lst==2)
 81	    {
 82		if (even%2==1&&i%2==0) continue;
 83		if (even%2==0&&i%2==1) continue;
 84	    }
 85	    cout<<"pos:"<<pos<<" i:"<<i<<endl;
 86	    res += dfs(pos-1,i%2==1?odd+1:0,i%2==0?even+1:0,i%2==0?2:1,limit&&i==mx,true);
 87	}
 88    }
 89    else
 90    {
 91	for ( int i = 0 ; i <= mx ; i++)
 92	{
 93	    cout<<"pos::"<<pos<<"  i:"<<i<<endl;
 94	    res += dfs(pos-1,i%2==1?1:0,i%2==0?1:0,0,limit&&i==mx,i==0?false:true);
 95	}
 96    }
 97	if (prehasnonzero&&!limit) dp[pos][odd][even][lst] = res;
 98    return res;
 99}  */
100LL solve(LL n)
101{
102    ms(digit,0);
103    int len = 0 ;
104    while (n)
105    {
106	digit[++len] = n;
107	n /= 10;
108    }
109	return dfs2(len,0,0,true,false);
110}
111int main()
112{
113	#ifndef  ONLINE_JUDGE 
114	freopen("code/in.txt","r",stdin);
115  #endif
116	int T;
117	cin>>T;
118	ms(dp2,-1);
119	int cas = 0 ;
120	while (T--)
121	{
122	    LL l,r;
123	    scanf("%lld%lld",&l,&r);
124	    LL ans = solve(r) - solve(l-1);
125	    printf("Case #%d: ",++cas);
126	    printf("%lld\n",ans);
127	}
128  #ifndef ONLINE_JUDGE  
129  fclose(stdin);
130  #endif
131    return 0;
132}