hdu 4734 F(x) （数位dp）
F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1140 Accepted Submission(s): 459
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
题意：
看懂f(x)函数之后，求[0,B]内的f(x)的值小于f(a)的有多少个。
思路：
数位dp，
由于f(x)的值不大，可以考虑把f(x)放入状态中。
dp[i][j][k] - i 位 j 开头f(x)值为k的数的个数。
那么有dp[i][j][k]=dp[i-1][p][k-j*2^(i-1)] 。
然后算一个区间内的个数时，一位一位往下算就够了。
考虑到时间的原因，采用了一个sum数组纪录前缀和。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 10005
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int n,m,ans,cnt,tot,flag;
int a,b,fac[10];
int dp[10][10][5500],sum[10][10][5500];
void solve()
{
int i,j,t=0,u,v;
for(i=9;i>=1;i--)
{
t+=((a/fac[i-1])%10)*(1<<i-1);
}
a=t;
ans=0;
for(i=9;i>=1;i--)
{
u=(b/fac[i-1])%10;
for(j=0;j<u;j++)
{
ans+=sum[i][j][a];
}
a-=u*(1<<i-1);
if(a<0) return ;
}
ans++;
}
int main()
{
int i,j,k,p,q,t,test=0;
fac[0]=1;
for(i=1;i<10;i++)
{
fac[i]=fac[i-1]*10;
}
dp[0][0][0]=1;
for(i=1;i<=9;i++)
{
for(j=0;j<=9;j++)
{
for(k=0;k<=5000;k++)
{
t=0;
for(p=0;p<=9;p++)
{
if((k-(1<<i-1)*j)>=0) t+=dp[i-1][p][k-(1<<i-1)*j];
}
dp[i][j][k]=t;
if(k>0) sum[i][j][k]=sum[i][j][k-1]+t;
else sum[i][j][k]=t;
}
}
}
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&a,&b);
solve();
printf("Case #%d: %d\n",++test,ans);
}
return 0;
}
/*
3
0 100
1 10
5 100
Case #1: 1
Case #2: 2
Case #3: 13
*/