【URAL 1056】Amount of Degree

Amount of Degrees
Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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Description
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
input output

15 20
2
2

3

Source
Problem Source: Rybinsk State Avia Academy

题意：求一个区间内满足条件的数有多少个，条件是该数可以写成k个b的不同次方的和

题解：建议看一下 刘聪的《浅谈数位类统计问题》那里很详细的写了，如何将该问题化成二进制的问题，然后用数位dp就能解决
下载地址：http://download.csdn.net/detail/qq_33583069/9569561

#include<stdio.h>
#include<string.h>
int k,b,all,num[108],dp[108][28];
int dfs(int hight,int cou,int limit)
{
    int res=0;
    if(cou<0) return 0;
    if(hight<=0) return cou==0?1:0;
    if(!limit&&~dp[hight][cou]) return dp[hight][cou];
    if(num[hight]||!limit) res+=dfs(hight-1,cou-1,limit);
    res+=dfs(hight-1,cou,limit&&!num[hight]);
    if(!limit) dp[hight][cou]=res;
    return res;
}
int solve(int n)
{
    int i;
    for(all=0;n;n=n/b) num[++all]=n%b;
    for(i=all;i>=1;i--) if(num[i]>1) break;
    for(;i>=1;i--) num[i]=1;

    return dfs(all,k,1);
}
int main()
{
    int x,y;

    memset(dp,-1,sizeof(dp));
    while(scanf("%d%d%d%d",&x,&y,&k,&b)>0)
    {
        printf("%d\n",solve(y)-solve(x-1));
    }

    return 0;
}