Timus Online Judge 1057. Amount of Degrees（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 input output 15 20 2 2 3

/*
题意: 求一个区间的 degree进制的1的个数为k的数的个数
思路:数位dp,一定要注意是1个个数为k  dp[i][j][k] 代表到达了i位的j进制还差k个1

详细注意的地方写在了代码中
*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8

typedef long long ll;

using namespace std;

#define N 35

int dp[33][15][33];

int degree,k;
int bit[N];

int dfs(int pos,int degree,int t,bool bound)
{
     if(t<0) return 0;
     if(pos==0) return t ? 0:1;
     if(!bound&&dp[pos][degree][t]>=0) return dp[pos][degree][t];
     int up=bound ? min(bit[pos],1):1;
     int ans=0;
     for(int i=0;i<=up;i++)
        ans+=dfs(pos-1,degree,t-i,bound&&i==bit[pos]); //必须是bit[pos]，不能是uo
     if(!bound) dp[pos][degree][t]=ans;
     return ans;
}

int solve(int x)
{
    int i,j;
    int len=0;
    while(x)
    {
        bit[++len]=x%degree;
        x/=degree;
    }
    return dfs(len,degree,k,true);
}

int main()
{
    int i,j,le,ri;
    memset(dp,-1,sizeof(dp));

    while(~scanf("%d%d",&le,&ri))
    {
        scanf("%d%d",&k,°ree);
        printf("%d\n",solve(ri)-solve(le-1));
    }
   return 0;
}