HDU 4734 F(x) 数位DP

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6398    Accepted Submission(s): 2465

Problem Description

For a decimal number x with n digits (A_nA_n-1A_n-2 ... A₂A₁), we define its weight as F(x) = A_n * 2^n-1 + A_n-1 * 2^n-2 + ... + A₂ * 2 + A₁ * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).

 

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
For each test case, there are two numbers A and B (0 <= A,B < 10⁹)

 

Output

For every case,you should output "Case #t: " at first, without quotes. The t is the case number starting from 1. Then output the answer.

 

Sample Input

30 1001 105 100

 

Sample Output

Case #1: 1Case #2: 2
Case #3: 13
#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>#include<vector>using namespace std;const int maxm = 10005;int dp[10][maxm], d[10], a, b, rev;int dfs(int len, int num, int flag){if (len < 0) return num >= 0;if (num < 0) return 0;if (!flag&&dp[len][num] != -1 && num != 1)return dp[len][num];int k = flag ? d[len] : 9;int ans = 0, next;for (int i = 0;i <= k;i++)ans += dfs(len - 1, num - i*(1 << len), flag&&i == k);if (!flag) dp[len][num] = ans;return ans;}int query(int n){int len = 0;while (n) d[len++] = n % 10, n /= 10;return dfs(len - 1, rev, 1);}int main(){int t, s;memset(dp, -1, sizeof(dp));scanf("%d", &t);for (int i = 1;i <= t;i++){scanf("%d%d", &a, &b);printf("Case #%d: ", i);rev = 0;s = 0;while (a){rev += (a % 10)*pow(2, s++);a /= 10;}printf("%d\n", query(b));}return 0;}/*30 100*/