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
",a)
#define ptlld(a) printf("%I64d
",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=4*atan(1.0);

using namespace std;
typedef long long ll;
template <class T> inline void read(T &n)
{
    char c; int flag = 1;
    for (c = getchar(); !(c >= '0' && c <= '9' || c == '-'); c = getchar()); if (c == '-') flag = -1, n = 0; else n = c - '0';
    for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0'; n *= flag;
}
ll Pow(ll base, ll n, ll mo)
{
    ll res=1;
    while(n)
    {
        if(n&1)
            res=res*base%mo;
        n>>=1;
        base=base*base%mo;
    }
    return res;
}
//***************************

int n;
const int maxn=100000+10;
const ll mod=1000000007;
int digit[30];
ll dp[30][2][2][2];
int tot=0;
ll dfs(int pos,int pre,int status,int ze,int limit)
{
    if(pos<0)
    {
        if(pre%2==1&&status%2==0)
            return 1;
        else if(pre%2==0&&status%2==1)
            return 1;
        else
            return 0;
    }
    if(!limit&&dp[pos][pre][status][ze]!=-1)
        return dp[pos][pre][status][ze];
    int ed=limit?digit[pos]:9;
    ll res=0;
    if(ze)
        res+=dfs(pos-1,0,0,1,limit&&(0==ed));
    else
    {
        if(pre%2==0)
            res+=dfs(pos-1,0,status^1,ze,limit&&(0==ed));
        else if(pre%2==1&&status==0)
            res+=dfs(pos-1,0,1,ze,limit&&(0==ed));
    }
    rep(i,1,ed)
    {
        if(i%2&&pre)
            res+=dfs(pos-1,1,status^1,0,limit&&(i==ed));
        else if(i%2==0&&!pre)
            res+=dfs(pos-1,0,status^1,0,limit&&(i==ed));
        else if(i%2&&!pre&&(status==1||ze))
            res+=dfs(pos-1,1,1,0,limit&&(i==ed));
        else if(i%2==0&&pre&&(status==0||ze))
            res+=dfs(pos-1,0,1,0,limit&&(i==ed));
    }
    if(!limit)
        dp[pos][pre][status][ze]=res;
    return res;
}
void work(ll num)
{
    tot=0;
    while(num)
    {
        digit[tot++]=num%10;
        num/=10;
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    int T_T;
    scanf("%d",&T_T);
    for(int kase=1;kase<=T_T;kase++)
    {
        printf("Case #%d: ",kase);
        ll num;
        sclld(num);
        work(num-1);
        ones(dp);
        ll temp=dfs(tot-1,0,0,1,1);
        sclld(num);
        work(num);
        ones(dp);
        ptlld(dfs(tot-1,0,0,1,1)-temp);
    }
    return 0;
}