HDU 5898 odd-even number 数位DP

odd-even number

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

 

　　

　　　　基础的数位DP

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 600+10, M = 1e2+11, mod = 1e9+7, inf = 0x3fffffff;

LL L,R,dp[80][2][80][2],vis[80][2][80][2];
int d[80];
long long dfs(int dep,int f,int p,int x) {
        if(dep < 0) return (p%2!=x%2);
        if(f && vis[dep][f][p][x]) return dp[dep][f][p][x];
        if(f) {
            long long& ret = dp[dep][f][p][x];
            vis[dep][f][p][x] = 1;
            for(int i=0;i<=9;++i) {
               if(x == -1) {
                if(i == 0)  ret += dfs(dep-1,f,1,-1);
                else ret += dfs(dep-1,f,1,i%2);
               }
               else {
                if(i%2 == x) {
                    ret += dfs(dep-1,f,p+1,i%2);
                }else {
                    if(p%2 == x) continue;
                    ret += dfs(dep-1,f,1,i%2);
                }
               }
            }
            return ret;

        } else {
            LL ret = 0;
            for(int i=0;i<=d[dep];++i) {
               if(x == -1) {
                if(i == 0)  ret += dfs(dep-1,i<d[dep],1,-1);
                else ret += dfs(dep-1,i<d[dep],1,i%2);
               }
               else {
                if(i%2 == x) {
                    ret += dfs(dep-1,i<d[dep],p+1,i%2);
                }else {
                    if(p%2 == x) continue;
                    ret += dfs(dep-1,i<d[dep],1,i%2);
                }
               }
            }
            return ret;
        }
}
long long solve(long long x)
{
    memset(dp,0,sizeof(dp));
    memset(vis,0,sizeof(vis));
    int len = 0;
    while(x){
        d[len++] = x%10;
        x/=10;
    }
    return dfs(len-1,0,0,-1);
}
int main() {
        int T,cas = 1;
        scanf("%d",&T);
        while(T--) {
            scanf("%I64d%I64d",&L,&R);
            printf("Case #%d: %I64d
",cas++,solve(R) - solve(L-1));
        }
        return 0;
}