HDu 4734【デジタルDP】
3556 ワード
F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6391 Accepted Submission(s): 2461
Problem Description
For a decimal number x with n digits (A
nA
n-1A
n-2 ... A
2A
1), we define its weight as F(x) = A
n * 2
n-1 + A
n-1 * 2
n-2 + ... + A
2 * 2 + A
1 * 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 < 10
9)
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
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6391 Accepted Submission(s): 2461
Problem Description
For a decimal number x with n digits (A
nA
n-1A
n-2 ... A
2A
1), we define its weight as F(x) = A
n * 2
n-1 + A
n-1 * 2
n-2 + ... + A
2 * 2 + A
1 * 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 < 10
9)
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
3 0 100 1 10 5 100
Sample Output
Case #1: 1 Case #2: 2 Case #3: 13
Source
题意理解：要求在不大于B的区间内找到不大于
F(A)的个数有多少个。
解题思路：这道题是一个数位DP的题，关键在于找到状态转移的条件。我们用dp[pos][sum]来保存结果，pos代表的是当前的位置，sum表示当前的依照F()的公式的和是多少，dp[pos][sum]即表示：长度为pos且权值不大于sum的数的个数。这样就可以让每个数所求的个数，是每个数的固有性质，不再需要每次查询都去memset(dp,-1,sizeof(dp)),如果每次都去memset，那必然超时，只需要一次就够了。
#include
#include
#include
using namespace std;
int a[20];
int dp[20][100010];
int totol;
int dfs(int pos,int sum,int limit)
{
if(pos==-1)
{
if(totol