hdu4734_数位dp

题意：

　　我们定义十进制数x的权值为f(x) = a(n)*2^(n-1)+a(n-1)*2(n-2)+...a(2)*2+a(1)*1，a(i)表示十进制数x中第i位的数字。

　　题目给出a，b，求出0~b有多少个不大于f(a)的数。

 ac代码

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <ctime>
 8 #include <queue>
 9 #include <list>
10 #include <set>
11 #include <map>
12 using namespace std;
13 #define INF 0x3f3f3f3f
14 typedef long long LL;
15 
16 int bit[12], dp[12][60000];
17 int t, A, B;
18 int solve(int len, int sum, int flag)
19 {
20     int res = 0;
21     if(sum < 0)
22         return 0;
23     if(len < 0)
24         return sum >= 0;
25     if(dp[len][sum] >= 0 && flag)
26         return dp[len][sum];
27     int te = flag ? 9 : bit[len];
28     for(int i = 0; i <= te; ++i)
29     {
30         int Sum = sum - i * (1 << len);
31         res += solve(len - 1, Sum, flag || i < te);
32     }
33     if(flag)
34         dp[len][sum] = res;
35     return res;
36 }
37 int main()
38 {
39     scanf("%d", &t);
40     memset(dp, -1, sizeof(dp));
41     for(int i = 1; i <= t; ++i)
42     {
43         scanf("%d %d", &A, &B);
44         int len = 0;
45         while(B)
46         {
47             bit[len++] = B % 10;
48             B /= 10;
49         }   
50         int temp = 0, j = 0;
51         while(A)
52         {
53             temp += (A % 10) * (1 << j);
54             j++;
55             A /= 10;
56         }
57         int res = solve(len-1, temp, 0);
58         printf("Case #%d: ", i);
59         printf("%d\n", res);
60     }
61     return 0;
62 }
View Code

wa代码，未解之谜

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <ctime>
 8 #include <queue>
 9 #include <list>
10 #include <set>
11 #include <map>
12 using namespace std;
13 #define INF 0x3f3f3f3f
14 typedef long long LL;
15 
16 int bit[12], dp[12][60000];
17 int t, A, B;
18 int solve(int len, int sum, int flag)
19 {
20     int res = 0;
21     if(sum > A)
22         return 0;
23     if(len < 0)
24         return sum <= A;
25     if(dp[len][sum] >= 0 && flag)
26         return dp[len][sum];
27     int te = flag ? 9 : bit[len];
28     for(int i = 0; i <= te; ++i)
29     {
30         int Sum = sum + i * (1 << len);
31         res += solve(len - 1, Sum, flag || i < te);
32     }
33     if(flag)
34         dp[len][sum] = res;
35     return res;
36 }
37 int main()
38 {
39     scanf("%d", &t);
40     memset(dp, -1, sizeof(dp));
41     for(int i = 1; i <= t; ++i)
42     {
43         scanf("%d %d", &A, &B);
44         int len = 0;
45         while(B)
46         {
47             bit[len++] = B % 10;
48             B /= 10;
49         }   
50         int temp = 0, j = 0;
51         while(A)
52         {
53             temp += (A % 10) * (1 << j);
54             j++;
55             A /= 10;
56         }
57         A = temp;
58         int res = solve(len-1, 0, 0);
59         printf("Case #%d: ", i);
60         printf("%d\n", res);
61     }
62     return 0;
63 }
View Code

 

posted @ 2016-08-22 14:44  海无泪  阅读(98)  评论(0)  编辑  收藏  举报