F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 382 Accepted Submission(s): 137
Problem Description
For a decimal number x with n digits (A
n
A
n-1
A
n-2
... A
2
A
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 )
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
meaning of the title
Define F(x) = A n * 2 n-1 + A n-1 * 2 n-2 + ... + A 2 * 2 + A 1 * 1. For A and B, find the number of F(x) from 0 to B that is less than or equal to F(A).
analyze
Digital DP. dp[i][j] is the number of digit i whose value is less than or equal to j. Use memoized search
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<algorithm> #include<cstring> #include<queue> #include<vector> #include<bitset> #include<map> #include<deque> #include<stack> using namespace std; typedef pair<int,int> pii; #define X first #define Y second #define pb push_back #define mp make_pair typedef long long ll; #define ms(a,b) memset(a,b,sizeof(a)) const int inf = 0x3f3f3f3f; const int maxn = 2e5; const int mod = 1e9+7; #define lson l,m,2*rt #define rson m+1,r,2*rt+1 int dp[15][maxn]; int bit[15]; int F(int A){ int len=0; int res=0; while(A){ res += (A%10)*(1<<len); len ++ ; A/=10; } return res; } int dfs(int pos,int num,bool limit){ if(pos==-1) return num>=0; if(num<0) return 0; if(!limit && dp[pos][num]!=-1) return dp[pos][num]; int ed=limit?bit[pos]:9; int ans=0; for(int i=0;i<=ed;i++){ ans+=dfs(pos-1,num-i*(1<<pos),limit && i==ed); } if(!limit) dp[pos][num]=ans; return ans; } int solve(int A,int B){ int maxx = F(A); int pos=0; while(B){ bit[pos++]=B%10; B/=10; } return dfs(pos-1,maxx,1); } int main(){ #ifdef LOCAL freopen("in.txt","r",stdin); #endif // LOCAL int t,A,B; int cas=1; scanf("%d",&t); ms(dp,-1); while(t--){ scanf("%d%d",&A,&B); printf("Case #%d: %d\n",cas++,solve(A,B)); } return 0; }