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 ... 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

30 1001 105 100

 

Sample Output

Case #1: 1Case #2: 2Case #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 99999999using 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^iint 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;}