Ural - 1057. Amount of Degrees

1057. Amount of Degrees

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 = 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≤231−11 ≤ X ≤ Y ≤ 2^31−1). The next two lines contain integers K and B (1≤K≤20;2≤B≤101 ≤ 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

思路：dp[cur][s][K]表示处理到当前状态，前面有s个不是1，一共需要K个不是1的，有多少

[code]#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=100;
int X,Y;
int K,B;
int dig[maxn];
int dp[maxn][20][25];
int dfs(int cur,int e,int s)
{
    if(cur<0)return s==K;
    if(s>K)return 0;
    if(!e&&dp[cur][s][K]!=-1)
        return dp[cur][s][K];
    int end=(e?min(1,dig[cur]):1);
    int ans=0;
    for(int i=0;i<=end;i++)
        ans+=dfs(cur-1,e&&i==dig[cur],s+(i!=0));//这里一定注意要是i==dig[cur]，因为前面的end是变化过的
    if(!e)dp[cur][s][K]=ans;
    return ans;
}
int solve(int x)
{
    int len=0;
    memset(dp,-1,sizeof(dp));
    while(x)
    {
        dig[len++]=x%B;
        x/=B;
    }
    return dfs(len-1,1,0);
}
int main()
{
    while(scanf("%d%d",&X,&Y)!=EOF)
    {
        scanf("%d%d",&K,&B);
        printf("%d\n",solve(Y)-solve(X-1));
    }
    return 0;
}