BNU52325-Increasing or Decreasing-数位DP-DFS

Posted 2020-08-14 HITLJR

题目地址：  https://www.bnuoj.com/v3/problem_show.php?pid=52325

两份代码，解释在第二份代码里面

第一份代码整理一下看着爽

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 long long  dp[20][11][3];
 6 int digit[20];
 7 int dfs(int len , int fore , int state , bool limit)
 8 {
 9     if (len == 0)
10         return 1;
11     if ( !limit && dp[len][fore][state] != -1 )
12         return dp[len][fore][state];
13     int res = 0;
14     int limitD = limit ? digit[len] : 9 ;
15     for (int j = 0 ; j<=limitD; j++)
16     {
17         int i;
18         if (fore == 10 && j == 0) i = 10;
19         else i = j;
20         if ( state == 0 ){
21             if ( fore == 10 || i == fore )
22                 res += dfs( len - 1 , i , 0 , limit && i == limitD );
23         }
24         else if ( state == 1 ){
25             if ( fore == 10 || i <= fore )
26                 res += dfs( len - 1 , i , 1 , limit && i == limitD );
27         }
28         else{
29             if ( fore == 10 || i >= fore )
30                 res += dfs( len - 1 , i , 2 , limit && i == limitD );
31         }
32     }
33     if (!limit)
34         dp[len][fore][state] = res;
35     return res;
36 }
37 long long solve(long long x)
38 {
39     int cnt = 0;
40     while (x)
41     {
42        digit[ ++cnt ] = ( int )( x % 10LL );
43        x /= 10LL;
44     }
45     return dfs( cnt , 10 , 1 , true ) + dfs( cnt , 10 , 2 , true ) - dfs( cnt , 10 , 0 , true );
46 }
47 int main()
48 {
49     memset(dp,-1,sizeof dp);
50     int N;
51     scanf("%d", &N );
52     long long  a,b;
53     while (N--)
54     {
55         scanf("%lld %lld" , &a , &b );
56         printf("%lld\\n", solve( b ) - solve ( a - 1LL ) );
57     }
58 }

View Code

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 //dp[i][j][k] 长度为i,前一位j,状态为k
 6 //k=0 相等 k=1 递增 k=2 递减
 7 long long  dp[20][11][3];
 8 int digit[20];
 9 //dfs计算在digit限制下，长度为len，前一位为fore，的state状态的数字，特别的令fore==10代表前面全是0
10 int dfs(int len , int fore , int state , bool limit)
11 {
12     //最边界计数，即枚举到最后一位，0即为全是前导0。
13     if (len == 0)
14     {
15         return 1;
16     }
17     //记忆化搜索
18     if (!limit && dp[len][fore][state] != -1)
19      return dp[len][fore][state];
20     int res = 0;
21     int limitD = limit?digit[len]:9;
22     for (int j = 0 ; j<=limitD; j++)
23     {
24         int i;
25         //确定填写的0,前导0变10,真0还是0
26         if (fore == 10 && j == 0) i = 10;
27         else i = j;
28         //如果前导0或者满足要求，那么继续dfs
29         //其中有些判断可以省略，但是所有的都加上fore==10,便于理解
30         if (state==0)      //相等
31         {
32                     if (fore==10||i==fore)
33                     res += dfs(len-1,i,0,limit&&i==limitD);
34         }
35         else if (state==1)//递减
36         {
37                     if (fore==10||i<=fore)
38                     res += dfs(len-1,i,1,limit&&i==limitD);
39          }
40          else           //递增
41          {
42                     if (fore==10||i>=fore)
43                     res += dfs(len-1,i,2,limit&&i==limitD);
44          }
45     }
46     if (!limit) dp[len][fore][state] = res;
47     return res;
48 }
49 long long solve(long long x)
50 {
51     int cnt = 0;
52     while (x)
53     {
54        digit[++cnt] = (int)(x % 10LL);
55        x /= 10LL;
56     }
57     //递增+递减-相等
58     return dfs(cnt,10,1,true) + dfs(cnt,10,2,true) - dfs(cnt,10,0,true);
59 }
60 int main()
61 {
62     memset(dp,-1,sizeof dp);
63     int N;
64     scanf("%d",&N);
65     long long  a,b;
66     while (N--)
67     {
68         scanf("%lld%lld",&a,&b);
69         printf("%lld\\n",solve(b) - solve(a-1LL) );
70     }
71 }