hdu 5898 odd-even number (数位dp)_bit even number-优快云博客

本文链接：https://blog.youkuaiyun.com/wchhlbt/article/details/52651323

本文介绍了一道关于寻找特定范围内奇偶数位交替出现的数字数量的问题，并通过数位DP的方法给出了详细的解答过程及代码实现。

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odd-even number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 661 Accepted Submission(s): 362

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

Sample Output

  
   Case #1: 29
Case #2: 36

Source

2016 ACM/ICPC Asia Regional Shenyang Online

解题思路：因为要求奇数连续出现偶数次，偶数连续出现奇数次，均是跟数位有关的操作，考虑使用数位dp思路。

采用和以前数位dp相同的思路，关于数位dp可以看http://blog.youkuaiyun.com/wchhlbt/article/details/52119930

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 20;
ll n;
ll dp[maxn][20][2], bit[maxn];


ll dfs(int pos,int pre,int limit,int ok)//pos数字位数，pre
{
    if(pos < 1 && ok) return 1;//如果顺利搜索到了第零位，说明这个数满足条件，返回1
    if(pos < 1 && !ok) return 0;
    if(!limit && dp[pos][pre][ok] != -1) return dp[pos][pre][ok];//记忆化搜索的关键
    ll ret = 0;
    int len = limit?bit[pos]:9;//如果处于非上界状态下，数字可以枚举0-9，处于上界状态则需要考虑给出的上界在该位的数字
    for(int i = 0; i <= len; i++)
    {
        if(pre==11) {
            if(i==0)
                ret += dfs(pos-1,  11, limit&&i==len, 1);
            else
                ret += dfs(pos-1, i, limit&&i==len, (i+1)%2);
            continue;
        }
        if(pre%2 != i%2)
        {
            if(ok==0)
                continue;
            else
            {
                ret += dfs(pos-1,  i, limit&&i==len, (i+1)%2);
            }
        }
        else
        {
            ret += dfs(pos-1,  i, limit&&i==len, !ok);
        }
        //ret += dfs(pos-1,  i, limit&&i==len);
    }
    if(!limit) dp[pos][pre][ok] = ret;
    return ret;
}


ll solve(ll n)//预处理函数，将所给定的数字，处理为每个位置上的数字（上界）
{
    int len = 0;
    while(n)
    {
        bit[++len] = n%10;
        n /= 10;
    }
    return dfs(len, 11, 1, 1);
}


int main()
{
    int t;
    ll a,b;
    cin>>t;
    memset(dp, -1, sizeof(dp));
    for(int i = 1; i<= t ; i++)
    {
        cin>>a>>b;
        //cout << solve(b) << ' ' << solve(a-1) << endl;
        printf("Case #%d: %I64d\n",i,solve(b)-solve(a-1));
    }
    return 0;
}