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 exactlyK 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 = 24+20,
18 = 24+21,
20 = 24+22.
Input
The first line of input contains integers X and Y, separated with a space (1 ≤ X ≤ Y ≤ 231−1). 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
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數位統計入門。
詳見程序備注（說白了就是填數字）
[cpp] #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; //考慮這位為1/0的情況
}
else if (a[i]==1)
{
ans+=C[i-1][k]; //考慮這位為0的情況
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;
}
#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; //考慮這位為1/0的情況
}
else if (a[i]==1)
{
ans+=C[i-1][k]; //考慮這位為0的情況
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;
}
