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HDU 4734 F(x)(數位DP)

編輯：關於C++

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 + 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 < 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

Sample Output

 Case #1: 1
Case #2: 2
Case #3: 13

簡單的數位DP：先將A表示成F(x)然後統計有多少大於x就行了。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pairpil;
const int INF = 0x3f3f3f3f;
int t,temp;
LL l,r;
int num[20];
LL dp[20][20000];
LL dfs(int pos,int sum,int flag)
{
    if(pos==0)
        return sum>=0;
    if(sum<0)  return 0;
    if(!flag&&dp[pos][sum]!=-1)
        return dp[pos][sum];
    LL ans=0;
    int ed=flag?num[pos]:9;
    for(int i=0;i<=ed;i++)
    {
        int s=sum-i*(1<<(pos-1));
        ans+=dfs(pos-1,s,flag&&i==ed);
    }
    if(!flag) dp[pos][sum]=ans;
    return ans;
}
LL solve(LL x)
{
    int pos=0;
    while(x)
    {
        num[++pos]=x%10;
        x/=10;
    }
    return dfs(pos,temp,1);
}
LL work(LL x)
{
    LL ans=0;
    int l=0;
    while(x)
    {
        ans+=(x%10)*(1<

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