HDU 2089 - 不要62

还是基础，不多说。。。- -坑爹悲催的我正在学数数。。。

 1 /*
 2 ID:esxgx1
 3 LANG:C++
 4 PROG:hdu2089
 5 */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 
12 #define NN    9
13 unsigned dp[NN][4];
14 
15 void work(int LN)
16 {
17     dp[0][0] = 1;
18     for(int i=1; i<=LN; ++i) {
19         dp[i][0] = dp[i-1][0] * 9 - dp[i-1][1];                        /// 无4和62的【含1】
20         dp[i][1] = dp[i-1][0];                                        /// 开头是2的, 无62无4
21         dp[i][2] = dp[i-1][2] * 10 + dp[i-1][0] + dp[i-1][1];        /// 有4或者62的
22 //        printf("%d 0=%d, 1=%d, 2=%d\n", i, dp[i][0], dp[i][1], dp[i][2]);
23     }
24 }
25 
26 unsigned solve(int i, int n, int &extra)
27 {
28     unsigned curr = n % 10;
29     unsigned rslt = curr * dp[i-1][2];
30     int extra0 = extra;
31     if (n/10) rslt += solve(i+1, n/10, extra0);
32     if (extra0) {
33         rslt += curr * dp[i-1][0];
34         extra = 1;
35     } else {
36         if (curr > 4) rslt += dp[i-1][0];
37         else if (curr == 4) extra = 1;
38 
39         if (n/10 % 10 > 6) rslt += dp[i][1];
40         else if (n/10 % 10 == 6) {
41             if (curr > 2) rslt += dp[i][1];
42             else if (curr == 2) extra = 1;
43         }
44     }
45     return rslt;
46 }
47 
48 #define sol(n)    ({int e=0; solve(1, n, e) + (e?1:0);})
49 
50 int main(void)
51 {
52     #ifndef ONLINE_JUDGE
53     freopen("in.txt", "r", stdin);
54     #endif
55     work(NN);
56     unsigned N, M;
57     while(scanf("%u%u", &N, &M) > 0) {
58         if (!(N || M)) break;
59         int e0, e1;
60         e0 = e1 = 0, N--;
61         printf("%u\n", M - N - (sol(M) - sol(N)));
62     }
63     return 0;
64 }

 

2014-08-07 23:43:32 Accepted 2089 0MS 320K 1357 B G++
posted @ 2014-08-07 23:47  e0e1e  阅读(146)  评论(0)  编辑  收藏  举报