Timus 1057 数位dp

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 2022

3

数位dp入门题，共有K种状态0~k-1分别表示有多少个B的M次幂在其中（i==2以及更大的情况直接排除）。

找到状态后，很容易做了。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<ctype.h>#include<algorithm>#include<string>#define PI acos(-1.0)#define maxn 40#define maxm 15#define INF 1<<25typedef long long ll;using namespace std;int ff[maxn][maxn];int bit[maxn];int aa,bb;int lim,bt;int solve (int pos,int p,bool cmp){    if(pos==0) return p==lim;    if(~ff[pos][p]&&!cmp) return ff[pos][p];    int u=cmp?bit[pos]:bt-1;    u=min(u,1);    int ans=0;    for(int i=0; i<=u; i++)    {        int newp;        if(i==1) newp=p+1;        else newp=p;        bool c=(cmp&&(i==bit[pos]));        if(newp>lim)        continue;        ans+=solve(pos-1,newp,c);    }    return cmp?ans:ff[pos][p]=ans;}int main(){    scanf("%d%d%d%d",&aa,&bb,&lim,&bt);    int pp=0;    aa--;    memset(ff,-1,sizeof(ff));    memset(bit,0,sizeof(bit));    while(aa)    {        bit[++pp]=aa%bt;        aa/=bt;    }    int l1=solve(pp,0,1);    pp=0;    memset(ff,-1,sizeof(ff));    memset(bit,0,sizeof(bit));    while(bb)    {        bit[++pp]=bb%bt;        bb/=bt;    }    int l2=solve(pp,0,1);    printf("%d\n",l2-l1);}