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HDU4734(数位dp)
F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4423 Accepted Submission(s): 1632
Problem Description
For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-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)
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
3
0 100
1 10
5 100
Sample Output
Case #1: 1
Case #2: 2
Case #3: 13
Source
2013 ACM/ICPC Asia Regional Chengdu Online
1 //2016.8.31 2 #pragma comment(linker, "/STACK:1024000000,1024000000") 3 #include <iostream> 4 #include <cstdio> 5 #include <cstring> 6 7 using namespace std; 8 9 int dp[20][10000], bit[20], a, b;//dp[i][j]表示第i位<=j的数目10 11 int F(int a)12 {13 int ans = 0, len = 0;14 while(a)15 {16 ans += (a%10)*(1<<len);17 len++;18 a /= 10;19 }20 return ans;21 }22 23 int dfs(int pos, int num, int fg)24 {25 if(pos == -1)return num>=0;26 if(num < 0)return 0;27 if(!fg && dp[pos][num]!=-1)//记忆化搜索28 return dp[pos][num];29 int ans = 0;30 int ed = fg?bit[pos]:9;31 for(int i = 0; i <= ed; i++)32 ans+=dfs(pos-1, num-i*(1<<pos), fg&&i==ed);33 if(!fg)dp[pos][num] = ans;34 return ans;35 }36 37 int solve(int b)38 {39 int len = 0;40 while(b)41 {42 bit[len++] = b%10;43 b /= 10;44 }45 int ans = dfs(len-1, F(a), 1);46 return ans;47 }48 49 int main()50 {51 int T, kase = 0;52 cin>>T;53 memset(dp, -1, sizeof(dp));54 while(T--)55 {56 scanf("%d%d", &a, &b);57 int ans = solve(b);58 printf("Case #%d: %d\n", ++kase, ans);59 }60 61 return 0;62 }
HDU4734(数位dp)
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