Ural 1057(Amount of Degrees-数位统计入门)

1057. Amount of Degrees Time limit: 1.0 second Memory limit: 64 MB Create a code to determine the amount of integers, lying in the set [ X; Y] and being a sum of exactly K different integer degrees of  B. Example. Let  X=15,  Y=20,  K=2,  B=2. By this example 3 numbers are the sum of exactly two integer degrees of number 2: 17 = 2 4+2 0, 18 = 2 4+2 1, 20 = 2 4+2 2. Input The first line of input contains integers  X and  Y, separated with a space (1 ≤  X ≤  Y ≤ 2 31−1). The next two lines contain integers  K and  B (1 ≤  K ≤ 20; 2 ≤  B ≤ 10). Output Output should contain a single integer — the amount of integers, lying between  X and  Y, being a sum of exactly  K different integer degrees of  B. Sample input output 15 20 2 2 3 Problem Source: Rybinsk State Avia Academy Tags: none   ( hide tags for unsolved problems ) Difficulty: 778     Printable version     Submit solution     Discussion (35) My submissions     All submissions (8220)     All accepted submissions (1664)    Solutions rating (1261)

数位统计入门。

详见程序备注（说白了就是填数字）

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MAXK (20+10)
#define MAXB (10+10)
#define MAXLog (32+10)
int x,y,k,b,C[MAXLog][MAXLog];
int a[MAXLog];
int calc(int x,int k,int b) //k num ()b
{
	int len=0,ans=0;
	while (x)
	{
		a[++len]=x%b;
		x/=b;
	}
	ForD(i,len)
	{
		if (a[i]>1)
		{
			ans+=C[i-1][k]+C[i-1][k-1];break; //考虑这位为1/0的情况 
		}
		else if (a[i]==1)
		{
			ans+=C[i-1][k]; //考虑这位为0的情况 
			k--;
		}
		if (k<0) return ans;
	}
	return ans;
}
int main()
{
//	freopen("ural1057.in","r",stdin);
	Rep(i,32+1)
	{
		C[i][0]=1;C[i][1]=i;
		Fork(j,2,i) C[i][j]=C[i-1][j]+C[i-1][j-1];
	}
//	cout<<C[3][1]<<' '<<C[3][2]<<' '<<C[3][3];
	scanf("%d%d%d%d",&x,&y,&k,&b);
	cout<<calc(y+1,k,b)-calc(x-1+1,k,b)<<endl;
	
	
	return 0;
}