hdu 5898 odd-even number 数位dp

odd-even number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).

 

Input

First line a t,then t cases.every line contains two integers L and R.

 

Output

Print the output for each case on one line in the format as shown below.

 

Sample Input

2 1 100 110 220

 

Sample Output

Case #1: 29 Case #2: 36

 

Source

2016 ACM/ICPC Asia Regional Shenyang Online

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e3+10,M=9e3+10,inf=2147483647,mod=1e9+7;
const ll INF=1e18+10;
ll f[20][3][20][3],bit[20];
ll dp(int pos,int pre,int now,int k,int flag,int p)
{
    //cout<<pos<<" "<<pre<<" "<<now<<" "<<k<<endl;
    if(pos==0)return (pre%2!=now%2&&k==0);
    if(flag&&f[pos][pre][now][k]!=-1)return f[pos][pre][now][k];
    int x=flag?9:bit[pos];
    ll ans=0;
    for(int i=0;i<=x;i++)
    {
        if(!p&&!i)
            ans+=dp(pos-1,1,0,0,flag||i<x,p||i);
        else
        {
            if(pre%2==i%2)
                ans+=dp(pos-1,i%2,now+1,k,flag||i<x,p||i);
            else
            {
                if(now%2!=pre%2)
                    ans+=dp(pos-1,i%2,1,k,flag||i<x,p||i);
                else
                    ans+=dp(pos-1,i%2,1,1,flag||i<x,p||i);
            }
        }
    }
    if(flag)f[pos][pre][now][k]=ans;
    return ans;
}
ll getans(ll x)
{
    int len=0;
    while(x)
    {
        bit[++len]=x%10;
        x/=10;
    }
    return dp(len,1,0,0,0,0);
}
int main()
{
    int T,cas=1;
    memset(f,-1,sizeof(f));
    scanf("%d",&T);
    while(T--)
    {
        ll l,r;
        scanf("%lld%lld",&l,&r);
        printf("Case #%d: %lld
",cas++,getans(r)-getans(l-1));
    }
    return 0;
}