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
#include
#include
#include
#include
#include
#include
#include
#include
#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;ik)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;
} 

你可能感兴趣的:(数位dp)