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ural 1057 Amount of Degrees(数位DP)

2015-12-18 12:43 399 查看


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

inputoutput
15 20
2
2

3

Problem Source: Rybinsk State Avia Academy

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题意：
首先定义了一种数为特殊数。

特殊数仅仅邮k个不同的b^x组成。

如今问你[x,y]里有多少这种特殊数。
思路：
数位DP。

dp[i][j]表示二进制表示中。前i为1的个数为j一共同拥有多少个数。其他进制类似处理。注意边界。
具体见代码：
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
typedef long long ll;
int dp[35][35],base[35],bin[35];
void init()
{
dp[0][0]=base[0]=1;
for(int i=1;i<=32;i++)
{
dp[i][0]=1;
base[i]=base[i-1]<<1;
for(int j=1;j<=i;j++)
dp[i][j]=dp[i-1][j]+dp[i-1][j-1];
}
}
int calc(int x,int k)
{
int i,one=0,ans=0;
for(i=31;i>=0;i--)
{
if(x&base[i])
{
if(one>k)
break;
ans+=dp[i][k-one];
one++;
x-=base[i];
}
}
if(one==k)
ans++;
return ans;
}
int getbin(int x,int b)
{
int ct=0,i,ret=0;
if(!x)
return x;
while(x)
{
bin[ct++]=x%b;
x/=b;
}
for(i=ct-1;i>=0;i--)
{
if(bin[i]>1)
break;
if(bin[i])
ret=ret<<1|1;
else
ret<<=1;
}
while(i>=0)
{
ret=ret<<1|1;
i--;
}
return ret;
}
int main()
{
int x,y,k,b;

init();
while(~scanf("%d%d%d%d",&x,&y,&k,&b))
{
y=getbin(y,b);
x=getbin(x-1,b);
printf("%d\n",calc(y,k)-calc(x,k));
}
return 0;
}
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