【hdu4734】F(x)-数位DP

题目描述：

For a decimal number x with n digits (A_nA_n-1A_n-2 ... A₂A₁), we define its weight as F(x) = A_n * 2^n-1 + A_n-1 * 2^n-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).

输入格式：

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)

输出格式：

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.

输入样例：

3

0 100

1 10

5 100

输出样例：

Case #1: 1

Case #2: 2

Case #3: 13 

思路：

设f[i] [j] [k] 表示第i位，首元素为j，当前计算的值为k 

转移方程f[i][j][l]+=f[i-1][k][l-(j<<(i-1))];

先预处理，由于多组数据，预处理一下前缀和，查询时先求和一下整数的部分，其余之后再求：

int i,j,p,di=1,ans=1 + (calc(n)<=m);
	for(i=1;b[i]<=n;i++)
	for(j=1;j<10;j++)
	ans+=f[i][j][m];
	for(;i;i--)
	{   
		p=n/b[i-1]%10;
		for(j=di;j<p;j++)ans+=f[i][j][m];
		m-=p<<(i-1),di=0;//求残余的部分，每次减去之前加的m
		if(m<0)break;
	}
	return ans;

　　

代码：

#include<cstdio>
#include<iostream>
#include<cstdlib>
using namespace std;
int f[10][10][10500],b[100];
int calc(int x)
{
	int t=1,tot=0;
	while(x) tot+=(x%10)*t,x/=10,t<<=1;
	return tot;
}
void init()
{
	int i,j,k,l;
	f[0][0][0]=b[0]=1;
	for(int i=1;i<10;i++)
	{
		b[i]=b[i-1]*10;
		for(j=0;j<10;j++)
		for(k=0;k<10;k++)
		for(l=j<<(i-1);l<=10000;l++)
		f[i][j][l]+=f[i-1][k][l-(j<<(i-1))];		
	}
	for( i=1;i<10;i++)
	for( j=0;j<10;j++)
	for( k=1;k<=10000;k++)
	f[i][j][k]+=f[i][j][k-1];
}
int sum(int n,int m)
{
	int i,j,p,di=1,ans=1 + (calc(n)<=m);
	for(i=1;b[i]<=n;i++)
	for(j=1;j<10;j++)
	ans+=f[i][j][m];
	for(;i;i--)
	{
		p=n/b[i-1]%10;
		for(j=di;j<p;j++)ans+=f[i][j][m];
		m-=p<<(i-1),di=0;
		if(m<0)break;
	}
	return ans;
}
int main()
{
	
	int T,i,a,bb;
	init();
	scanf("%d",&T);
	for(i=1;i<=T;i++){
		scanf("%d%d",&a,&bb);
		printf("Case #%d: %d
",i,sum(bb,calc(a)));
	}
	return 0;
}