Hdu-4734 F(x)(数位DP)

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. Thet 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

分析：这里有一个小优化技巧，dp[i][j]表示目前第i位，最多再凑j的方案数，这样一来dp数组的值就和n无关了。

#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<ctime>  #include<cstring>#include<stack>#include<cmath>#include<queue>#define INF 0x3f3f3f3f#define eps 1e-9#define MAXN 100000using namespace std;int T,n,m,dp[12][10000],a[12];int dfs(int pos,int sta,int limit){if(pos == -1) return sta <= n;if(sta > n) return 0;if(!limit && dp[pos][n-sta] >= 0) return dp[pos][n-sta];int up = limit ? a[pos] : 9;int ans = 0;for(int i = 0;i <= up;i++){ans += dfs(pos-1,sta + i*(1<<pos),limit && i == up);}if(!limit) dp[pos][n-sta] = ans;return ans;}int got(int x){int ans = 0,pos = 0;while(x){ans += (x % 10)*(1<<pos);x /= 10;pos++;}return ans;}int solve(int x){int pos = 0;while(x){a[pos++] = x % 10;x /= 10;}return dfs(pos-1,0,1);}int main(){memset(dp,-1,sizeof(dp));scanf("%d",&T);for(int t = 1;t <= T;t++){scanf("%d%d",&n,&m);n = got(n);printf("Case #%d: %d\n",t,solve(m));}}