hdu 5898 odd-even number 数位DP

传送门：hdu 5898 odd-even number

思路：数位DP，套着数位DP的模板搞一发就可以了不过要注意前导0的处理，dp[pos][pre][status][ze]

pos:当前处理的位

pre:上一位的奇偶性

status：截止到上一位的连续段的奇偶性

ze:是否有前导0

/**************************************************************
    Problem:hdu 5898 odd-even number
    User: youmi
    Language: C++
    Result: Accepted
    Time:0MS
    Memory:1580K
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#define zeros(a) memset(a,0,sizeof(a))
#define ones(a) memset(a,-1,sizeof(a))
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define sclld(a) scanf("%I64d",&a)
#define pt(a) printf("%d\n",a)
#define ptlld(a) printf("%I64d\n",a)
#define rep(i,from,to) for(int i=from;i<=to;i++)
#define irep(i,to,from) for(int i=to;i>=from;i--)
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define lson (step<<1)
#define rson (lson+1)
#define eps 1e-6
#define oo 0x3fffffff
#define TEST cout<<"*************************"<<endl
const double pi=*atan(1.0);

using namespace std;
typedef long long ll;
template <class T> inline void read(T &n)
{
    char c; int flag = ;
    for (c = getchar(); !(c >= '' && c <= '' || c == '-'); c = getchar()); if (c == '-') flag = -, n = ; else n = c - '';
    for (c = getchar(); c >= '' && c <= ''; c = getchar()) n = n *  + c - ''; n *= flag;
}
ll Pow(ll base, ll n, ll mo)
{
    ll res=;
    while(n)
    {
        if(n&)
            res=res*base%mo;
        n>>=;
        base=base*base%mo;
    }
    return res;
}
//***************************

int n;
const int maxn=+;
const ll mod=;
int digit[];
ll dp[][][][];
int tot=;
ll dfs(int pos,int pre,int status,int ze,int limit)
{
    if(pos<)
    {
        if(pre%==&&status%==)
            return ;
        else if(pre%==&&status%==)
            return ;
        else
            return ;
    }
    if(!limit&&dp[pos][pre][status][ze]!=-)
        return dp[pos][pre][status][ze];
    int ed=limit?digit[pos]:;
    ll res=;
    if(ze)
        res+=dfs(pos-,,,,limit&&(==ed));
    else
    {
        if(pre%==)
            res+=dfs(pos-,,status^,ze,limit&&(==ed));
        else if(pre%==&&status==)
            res+=dfs(pos-,,,ze,limit&&(==ed));
    }
    rep(i,,ed)
    {
        if(i%&&pre)
            res+=dfs(pos-,,status^,,limit&&(i==ed));
        else if(i%==&&!pre)
            res+=dfs(pos-,,status^,,limit&&(i==ed));
        else if(i%&&!pre&&(status==||ze))
            res+=dfs(pos-,,,,limit&&(i==ed));
        else if(i%==&&pre&&(status==||ze))
            res+=dfs(pos-,,,,limit&&(i==ed));
    }
    if(!limit)
        dp[pos][pre][status][ze]=res;
    return res;
}
void work(ll num)
{
    tot=;
    while(num)
    {
        digit[tot++]=num%;
        num/=;
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    int T_T;
    scanf("%d",&T_T);
    for(int kase=;kase<=T_T;kase++)
    {
        printf("Case #%d: ",kase);
        ll num;
        sclld(num);
        work(num-);
        ones(dp);
        ll temp=dfs(tot-,,,,);
        sclld(num);
        work(num);
        ones(dp);
        ptlld(dfs(tot-,,,,)-temp);
    }
    return ;
}