hdu4734之数位dp

F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 606    Accepted Submission(s): 236

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

2013 ACM/ICPC Asia Regional Chengdu Online

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;

const int MAX=10;
int dp[MAX][5000];//表示剩余i位最大可加j时的方法数
int a,b,pow[MAX][MAX];//pow表示j*2^i
int digit[MAX];

void Init(){
	memset(dp,-1,sizeof dp);
	for(int i=1;i<MAX;++i)pow[i][0]=i;
	for(int i=1;i<MAX;++i){
		for(int j=1;j<MAX;++j)pow[j][i]=pow[j][i-1]*2;
	}
}

int digit_dp(int k,int size,int flag){
	if(!flag && dp[size][k] != -1)return dp[size][k];
	if(!size)return 1;
	int m=flag?digit[size]:9,sum=0;
	for(int i=0;i<=m;++i){
		if(pow[i][size-1]>k)break;
		sum+=digit_dp(k-pow[i][size-1],size-1,flag && i == digit[size]);
	}
	if(!flag)dp[size][k]=sum;
	return sum;
}

int calculate(int &b,int &a){
	int n=0,size=0;
	while(a)n+=pow[a%10][size++],a/=10;
	a=n,size=0;
	while(b)digit[++size]=b%10,b/=10;
	return digit_dp(a,size,1);
} 

int main(){
	Init();
	int t,num=0;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&a,&b);
		printf("Case #%d: %d\n",++num,calculate(b,a));
	} 
	return 0;
}