F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3530 Accepted Submission(s): 1317
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)
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
题意：找出i在0到b之间的f(i)小于等于f(a)的数的个数。
思路：数位dp。主要在于状态转移不好想。dp[i][j]表示i位数比j小的数的个数。用递归完成的话就只需要思考边界和状态转移。
边界：
dp[i][j]如果j小于0，显然是dp[i][j]=0的，如果i==0，说明就是0，显然任何数都比0大，所以dp[i][j]对于j>=0的时候dp[i][j]=1,否则dp[i][j]=0。
状态转移：
dp[i][j]+=dp[i-1][j-k*(1<<(i-1))]；
完成上述两步推导就能开始写这题了。
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
int dp[20][9300];
int digit[20];
int dfs(int pos,int st,bool limit)
{
if(pos==0)return st>=0;
if(st<0)return 0;
if(!limit&&dp[pos][st]!=-1)return dp[pos][st];
int ans=0;
int end=limit?digit[pos]:9;
for(int i=0; i<=end; i++)
ans+=dfs(pos-1,st-i*(1<<(pos-1)),limit&&(i==end));
if(!limit)
dp[pos][st]=ans;
return ans;
}
int f(int x)
{
int ans=0;
int i=0;
while(x)
ans+=(x%10)*(1<<(i++)),x/=10;
return ans;
}
int get(int a,int b)
{
int bj=0;
while(b)
digit[++bj]=b%10,b/=10;
return dfs(bj,f(a),1);
}
int main()
{
int t,o=1;
memset(dp,-1,sizeof(dp));
scanf("%d",&t);
while(t--)
{
int a,b;
scanf("%d%d",&a,&b);
printf("Case #%d: %d\n",o++,get(a,b));
}
return 0;
}