找规律/数位DP HDOJ 4722 Good Numbers

 

题目传送门

 1 /*
 2     找规律/数位DP：我做的时候差一点做出来了，只是不知道最后的 is_one ()
 3                     http://www.cnblogs.com/crazyapple/p/3315436.html
 4     数位DP：http://blog.csdn.net/libin56842/article/details/11580497
 5 */
 6 #include <cstdio>
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <cstring>
10 #include <string>
11 #include <cmath>
12 using namespace std;
13 
14 const int MAXN = 1e5 + 10;
15 const int INF = 0x3f3f3f3f;
16 
17 int is_one(long long n)
18 {
19     long long m = n;
20     long long i = n / 10 * 10;
21 
22     for (; i<=m; ++i)
23     {
24         long long tmp = i;    int sum = 0;
25         while (tmp)
26         {
27             sum += tmp % 10;
28             tmp /= 10;
29         }
30         if (sum % 10 == 0)    return 1;
31     }
32 
33     return 0;
34 }
35 
36 long long get_num(long long n)
37 {
38     if (n < 0)    return 0;
39     if (n <= 10)    return 1;
40 
41     return n/10 + is_one (n);
42 }
43 
44 int main(void)        //HDOJ 4722 Good Numbers
45 {
46     //freopen ("G.in", "r", stdin);
47 
48     int t, cas = 0;
49     long long l, r;
50     scanf ("%d", &t);
51     while (t--)
52     {
53         scanf ("%I64d%I64d", &l, &r);
54 
55         printf ("Case #%d: %I64d\n", ++cas, get_num (r) - get_num (l-1));
56     }
57 
58     return 0;
59 }
60 
61 
62 /*
63 Case #1: 0
64 Case #2: 1
65 Case #3: 1
66 */
 1 /*
 2     数位DP：基础题，对于一个数，把它每位数分出来，从最高位开始枚举。dp[i][(j+k)%10] += dp[i+1][j];
 3             dp[i][j]表示前i位数和取模是j的个数
 4 */
 5 #include <cstdio>
 6 #include <algorithm>
 7 #include <cstring>
 8 #include <cmath>
 9 using namespace std;
10 
11 typedef long long ll;
12 const int MAXN = 22;
13 const int INF = 0x3f3f3f3f;
14 ll dp[MAXN][MAXN];
15 int dig[MAXN];
16 
17 ll work(ll n)   {
18     ll tmp = n; int len = 0;
19     memset (dig, 0, sizeof (dig));
20     while (tmp) {
21         dig[++len] = tmp % 10;
22         tmp /= 10;
23     }
24     memset (dp, 0, sizeof (dp));    int x = 0;
25     for (int i=len; i>=1; --i)  {
26         for (int j=0; j<10; ++j)    {
27             for (int k=0; k<10; ++k)    {
28                 dp[i][(j+k)%10] += dp[i+1][j];
29             }
30         }
31         for (int j=0; j<dig[i]; ++j)    dp[i][(x+j)%10]++;
32         x = (x + dig[i]) % 10;
33     }
34     if (!x) dp[1][0]++;
35     return dp[1][0];
36 }
37 
38 int main(void)  {       //HDOJ 4722 Good Numbers
39     int T, cas = 0;  scanf ("%d", &T);
40     while (T--) {
41         ll l, r;    scanf ("%I64d%I64d", &l, &r);
42         printf ("Case #%d: %I64d\n", ++cas, work (r) - work (l-1));
43     }
44 
45     return 0;
46 }
数位DP

 

posted @ 2015-04-19 20:36  Running_Time  阅读(233)  评论(0)    收藏  举报