hdu 4734 F(x) 数位dp

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

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

数位dp；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
ll f[20][N],bit[20];
ll dp(int pos,int pre,int flag)
{
    if(pre<0)return 0;
    if(pos==0)return (pre>=0);
    if(flag&&f[pos][pre]!=-1)return f[pos][pre];
    ll x=flag?9:bit[pos];
    ll ans=0;
    for(ll i=0;i<=x;i++)
    {
        ans+=dp(pos-1,pre-i*(1<<(pos-1)),flag||i<x);
    }
    if(flag)f[pos][pre]=ans;
    return ans;
}
int check(ll x)
{
    int sum=0,base=1;
    for(int i=0;x;i++,base*=2,x/=10)
        sum+=(x%10)*base;
    return sum;
}
ll getans(ll x,ll y)
{
    int len=0;
    while(x)
    {
        bit[++len]=x%10;
        x/=10;
    }
    return dp(len,check(y),0);
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    memset(f,-1,sizeof(f));
    while(T--)
    {
        ll l,r;
        scanf("%lld%lld",&l,&r);
        //cout<<check(l)<<endl;
        printf("Case #%d: %lld
",cas++,getans(r,l));
    }
    return 0;
}