<span>hdu5898 odd-even number（数位dp）</span>

题意：

求L到R区间内，连续奇数个数是偶数，连续偶数个数是奇数的数的个数

思路：

裸数位dp，赛场上忘了不合法的break，妈的调了一个多小时简直是日了狗了！

本来就是蒟蒻还感冒了什么题都写不出来

/* ***********************************************
Author        :devil
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <string>
#include <cmath>
#include <stdlib.h>
#define inf 0x3f3f3f3f
#define LL long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ou(a) printf("%d\n",a)
#define pb push_back
#define mkp make_pair
template<class T>inline void rd(T &x){char c=getchar();x=0;while(!isdigit(c))c=getchar();while(isdigit(c)){x=x*10+c-'0';c=getchar();}}
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;
const int mod=1e9+7;
const int N=310;
LL dp[22][4];
void init()
{
    dp[1][0]=1;
    dp[1][2]=1;
    for(int i=2;i<20;i++)
    {
        dp[i][0]=(dp[i-1][2]+dp[i-1][1])*5;
        dp[i][1]=dp[i-1][0]*5;
        dp[i][2]=(dp[i-1][1]+dp[i-1][3])*5;
        dp[i][3]=dp[i-1][2]*5;
    }
}
int bit[21];
LL solve(LL n)
{
    int len=0;
    while(n)
    {
        bit[++len]=n%10;
        n/=10;
    }
    bit[len+1]=0;
    LL ans=0;
    int tp=bit[len]/2;
    ans=ans+dp[len][1]*tp;
    if(bit[len]%2==0) tp--;
    ans=ans+dp[len][2]*tp;
    for(int i=len-1;i>=1;i--)
    {
        ans=ans+dp[i][1]*5;
        ans=ans+dp[i][2]*4;
    }
    int now=bit[len]%2,ln=1;
    for(int i=len-1;i>=1;i--)
    {
        if(now%2==1)
        {
            if(ln%2==1)
            {
                ans=ans+dp[i][0]*(bit[i]/2);
                if(bit[i]%2) ln++;
                else break;
            }
            else
            {
                int tp=bit[i]/2;
                ans=ans+dp[i][1]*tp;
                if(bit[i]%2!=0) tp++;
                ans=ans+dp[i][2]*tp;
                if(bit[i]%2) ln++;
                else
                {
                    now=0;
                    ln=1;
                }
            }
        }
        else
        {
            if(ln%2==1)
            {
                int tp=bit[i]/2;
                ans=ans+dp[i][1]*tp;
                if(bit[i]%2) tp++;
                ans=ans+dp[i][3]*tp;
                if(bit[i]%2)
                {
                    now=1;
                    ln=1;
                }
                else ln++;
            }
            else
            {
                int tp=bit[i]/2;
                if(bit[i]%2!=0) tp++;
                ans=ans+dp[i][2]*tp;
                if(bit[i]%2) break;
                else ln++;
            }
        }
    }
    return ans;
}
int main()
{
    init();
    int t,cas=0;
    LL n,m;
    rd(t);
    while(t--)
    {
        scanf("%lld%lld",&n,&m);
        printf("Case #%d: %lld\n",++cas,solve(m+1)-solve(n));
    }
    return 0;
}