ural1057 Amount of Degrees 【数位dp】论文例题

URAL - 1057

Amount of Degrees

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

SubmitStatus

Description

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 ofB.

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 andY, separated with a space (1 ≤  X ≤  Y ≤ 2 ³¹−1). The next two lines contain integersK and B (1 ≤  K ≤ 20; 2 ≤  B ≤ 10).

Output

Output should contain a single integer — the amount of integers, lying betweenX and Y, being a sum of exactly K different integer degrees ofB.

Sample Input

inputoutput

15 2022

3

Source

Problem Source: Rybinsk State Avia Academy

其实是做完这个题才做的nefu的题的，论文讲的太粗狂了，看了标称才写对==我上午超级不理解的地方是为什么当发现当前位>1时，我们加的是f[i+1][k-tot]？为啥这货就代表全都选啊啊啊。我们想想f[i][j]的递推表达式是f[i][j]=f[i-1][j]+f[i-1][j-1]前者表示左子树的根节点是0，后者表示右子树的根节点是1，所以f[i+1][j]就表示长度为i时的所以情况啦！

/***********ural10572016.2.163961msG++ 4.9***********/#include <iostream>#include<cstdio>#include<cstring>#include<vector>using namespace std;int f[32][32];void init(){    f[0][0]=1;    for(int i=1;i<=31;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 cal(int n,int k,int b){    vector<int>c;    int ans=0,tot=0;    while(n)    {        c.push_back(n%b);        n/=b;    }    for(int i=c.size()-1;i>=0;i--)    {        if(c[i]==1)        {            ans+=f[i][k-tot];            tot++;            if(tot>=k) break;        }        if(c[i]>1)        {            ans+=f[i+1][k-tot];            break;        }    }    if(tot==k) ans++;    return ans;}int main(){  // freopen("cin.txt","r",stdin);    init();    int x,y,b,k;   // printf("15,4 %d\n",change(15,4));    while(~scanf("%d%d",&x,&y))    {        scanf("%d%d",&k,&b);        printf("%d\n",cal(y,k,b)-cal(x-1,k,b));    }    return 0;}