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hdu 4734 解題報告

題目:

F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9119 Accepted Submission(s): 3607

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)

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


程式碼:

/* hdu-4734 */
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int N=1e5+5;
int digits[10];
int dp[10][N];
int
a,b; /* 引數:' dp[pos][pre] - pre最初等於f(a),比如pos為是3,那麼執行下一個pos位時,pre就要減去3*(1<<pos), pos - 位置 limit - 判斷是否是最大值,為真表示是最大值 */ int f(int x) { if(x==0) return 0; int ans=f(x/10); return ans*2+(x%10); } int dfs(int pos,int pre, int limit) { if(pos==-1) return pre>=0; if(pre<0 ) return 0; if(!limit&&dp[pos][pre]) return dp[pos][pre]; int maxd=limit?digits[pos]:9; int ans=0;//計數 for(int i=0;i<=maxd;i++) { ans+=dfs(pos-1,pre-i*(1<<pos),limit&&i==digits[pos]); } if(!limit) dp[pos][pre]=ans; return ans; } int solve(int n) { int len=0; while(n) { digits[len++]=n%10; n/=10; } return dfs(len-1,f(a),true); } int main() { int t; scanf("%d",&t); for(int i=1;i<=t;i++) { scanf("%d%d",&a,&b); printf("Case #%d: %d\n",i,solve(b)); } return 0; }