URAL 1057 (数位dp)

问题描述:

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 Xand Y, being a sum of exactly K different integer degrees of B.

Example

15 20

2

2

3

题目题意:题目给我们了一个区间，问这个区间内有多少个数可以被表示成B^(a1)+B^(a2)+....B^(ak) 且（a1!=a2!=...!=ak)这种形式.

题目分析:先看一下这个分析报告吧！点击打开链接

然后看看这个大神的博客点击打开链接

鶸鸡努力中

代码如下:

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;const int maxn=35;int f[maxn][maxn];int X,Y,k,b;void init(){    f[0][0]=1;    for (int i=1;i<=32;i++) {        f[i][0]=f[i-1][0];        for (int j=1;j<=i;j++) {            f[i][j]=f[i-1][j]+f[i-1][j-1];        }    }}int solve(int n){    int a[100],len=0,cnt=0,ans=0;    while (n) {        a[len++]=n%b;        n=n/b;    }    for (int i=len-1;i>=0;i--) {        if (a[i]==1) {            ans+=f[i][k-cnt];            cnt++;            if (cnt==k) break;        }        else if (a[i]>1){            ans+=f[i+1][k-cnt];            break;        }    }    if (cnt==k) ans++;    return ans;}int main(){    init();    while (scanf("%d%d%d%d",&X,&Y,&k,&b)!=EOF) {        printf("%d\n",solve(Y)-solve(X-1));    }    return 0;}