Hdu 4734 F(x) （数位dp）
题目链接
F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2272 Accepted Submission(s): 861
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
Source
2013 ACM/ICPC Asia Regional Chengdu Online
题解:F(x)的最大值不超过5000，可以用数位dp来做。
用dp[i][j]表示0到i位还没有填（即不超过i位的所有数字中），要让F值小于等于等于j的方案数。
dp[i][j]+=dp[i-1][j-x*(1<<i)], 0<=x<=9;
数位统计问题的具体统计方法可以参加相关数位dp的论文。
代码如下：
#include<cstdio>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<iostream>
#include<map>
#include<vector>
#include<algorithm>
#include<set>
#include<cmath>
using namespace std;
const int nn = 3100;
const int inf = 0x3fffffff;
int A,B;
int wei[20];
int dp[20][5000];
int F(int x)
{
int re=0;
int ix=0;
while(x)
{
if(x%10)
re+=x%10*(1<<ix);
x/=10;
ix++;
}
return re;
}
int dfs(int id,int lim,bool man)
{
if(lim<0)
return 0;
if(id==-1)
{
return 1;
}
if(!man&&dp[id][lim]!=-1)
return dp[id][lim];
int re=0;
int en=man?wei[id]:9;
for(int i=0;i<=en;i++)
{
re+=dfs(id-1,lim-i*(1<<(id)),man&&(i==en));
}
if(!man)
dp[id][lim]=re;
return re;
}
int solve()
{
int ix=0;
memset(wei,0,sizeof(wei));
while(B)
{
if(B%10)
{
wei[ix]=B%10;
}
ix++;
B/=10;
}
return dfs(10,F(A),true);
}
int main()
{
memset(dp,-1,sizeof(dp));
int t,cas=1;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&A,&B);
printf("Case #%d: %d\n",cas++,solve());
}
return 0;
}