F(x)
http://acm.hdu.edu.cn/showproblem.php?pid=4734
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9614 Accepted Submission(s): 3802
Problem Description
For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-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)
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 13000005 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef pair<int,int> pii; 14 typedef pair<long long,int>pli; 15 typedef pair<char,int> pci; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long MOD=1e9+7; 19 /*#ifndef ONLINE_JUDGE 20 freopen("1.txt","r",stdin); 21 #endif */ 22 23 int dp[25][10005]; 24 int a[25]; 25 int num; 26 27 int fun(int x){ 28 if(!x) return 0; 29 int ans=fun(x/10); 30 return ans*2+(x%10); 31 } 32 33 int dfs(int pos,int sum,int limit){ 34 if(pos==-1) return sum<=num; 35 if(sum>num) return 0; 36 if(!limit&&dp[pos][num-sum]!=-1) return dp[pos][num-sum]; 37 int up = limit ? a[pos] : 9; 38 int ans=0; 39 for(int i=0;i<=up;i++){ 40 ans+=dfs(pos-1,sum+i*(1<<pos),limit&&i==a[pos]); 41 } 42 if(!limit) dp[pos][num-sum]=ans; 43 return ans; 44 } 45 int solve(int x){ 46 int pos=0; 47 while(x){ 48 a[pos++]=x%10; 49 x/=10; 50 } 51 return dfs(pos-1,0,true); 52 } 53 54 int main(){ 55 #ifndef ONLINE_JUDGE 56 // freopen("1.txt","r",stdin); 57 #endif 58 std::ios::sync_with_stdio(false); 59 int n,m; 60 int t; 61 cin>>t; 62 memset(dp,-1,sizeof(dp)); 63 for(int i=1;i<=t;i++){ 64 cin>>n>>m; 65 num=fun(n); 66 int ans=solve(m); 67 cout<<"Case #"<<i<<": "<<ans<<endl; 68 } 69 }