寒假集训日志——ＤＰ

Posted 2020-06-09

（一）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 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 ≤ 2³¹−1). 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

一道据说比较经典的题目，基本弄懂，ans++ 那里实在没搞懂，不过也没那么多时间给我犹豫。。

主要借助一个状态转移方程 f[i][j] = f[i-1][j-1] + f[i-1][j]

f[i][j] 表示二进制中前ｉ位中有ｊ个１的个数，其他进制的数在转换过来之后都可一借用f[i][j]来解决。可以借助这个图理解一下这个方程。

不过ａｎｓ＋＋那里实在每看懂。

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<fstream>
#include<vector>
typedef long long LL;
using namespace std;

const int maxN = 50;
int f[maxN][maxN];

void getMap (){
    f[0][ 0] = 1;
    for( int i = 1 ; i< maxN ;++i){
        f[i][0] = 1;
        for( int j=  1 ; j<=i ;++j)
            f[i][j] = f[i-1][j] + f[i-1][j-1] ;
    }
}

int solve ( int x, int k ,  int c) {
    vector< int > arr;
    while( x) {
        arr.push_back ( x% c);
        x /= c;
    }
    int ans = 0, cnt = 0  ;
    for( int i = arr.size() -1 ; i>= 0 ; --i){
        if( arr[i] == 1){
            ans += f[i][ k - cnt];
            cnt++ ;
            if( cnt == k )  break;
        }
        else if( arr[i] > 1){
            ans += f[i+1][k - cnt];
            break;
        }
    }
    if( cnt == k )  ans ++;
    return ans ;
}

int main()
{
    for( int i = 0 ; i< maxN ; ++i)
        for( int j = 0 ; j< maxN; ++j)
                f[i][j] =  0 ;
    int x, y ;
    int k, b;
    getMap();
    cin>>x>>y>>k>>b;
    cout<<solve(y,k,b) - solve(x-1 , k , b)<<endl;
return 0;
}

 

（二）树形DP