zoukankan      html  css  js  c++  java
  • hdu 6093---Rikka with Number(计数)

    题目链接

    Problem Description
    As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

    In radix d, a number K=(A1A2...Am)d(Ai[0,d),A10) is good if and only A1Am is a permutation of numbers from 0 to d1.

    A number K is good if and only if there exists at least one d2 and K is good under radix d.

    Now, Yuta wants to calculate the number of good numbers in interval [L,R]

    It is too difficult for Rikka. Can you help her?  
     
    Input
    The first line contains a number t(1t20), the number of the testcases. 

    For each testcase, the first line contains two decimal numbers L,R(1LR105000).
     
    Output
    For each testcase, print a single line with a single number -- the answer modulo 998244353.
     
    Sample Input
    2
    5 20
    123456 123456789
     
    Sample Output
    3
    114480
     
     
    题意: 一个数 x 如果用 d 进制进行表示,是0~d-1的一个排列(不能以0打头),那么这个数称为 “优数”,一个数只要存在任意一个 d 进制符合0~d-1的一个排列,那么就是“优数”,问L到R区间有多少个优数?
     
    思路:按进制计算,找边界,一定存在dl和dr是边界,dl+1~dr-1 进制的所有排列都在L~R之间。
           
    官方题解:
                   
     
    代码如下:
    import java.math.BigInteger;
    import java.util.Scanner;
    
    public class Main {
        static int MAXN = 1600;
        static BigInteger[] dx = new BigInteger[MAXN];
        static long MOD = 998244353;
        static long [] fac = new long [MAXN];
        static void Init(){
            dx[0] = BigInteger.ZERO;
            dx[1] = BigInteger.ZERO;
            for(int i=2; i<MAXN; i++){
                dx[i] = BigInteger.ZERO;
                for(int j=i-1; j>=0; j--){
                    dx[i] = dx[i].multiply(BigInteger.valueOf(i)).add(BigInteger.valueOf(j));
                }
            }
            fac[0] = fac[1] = 1;
            for(int i=2; i<MAXN; i++) fac[i]=fac[i-1]*i%MOD;
        }
        static int Low(BigInteger x){
            int low=0, high=MAXN-1;
            while(low < high){
                int mid = (low+high)/2;
                if(dx[mid].compareTo(x)>=0)high = mid;
                else low = mid+1;
            }
            return high;
        }
        public static void main(String[] args){
            Init();
            Scanner in = new Scanner(System.in);
            int T = in.nextInt();
            for(int cas=1; cas<=T; cas++){
                int [] vis = new int [MAXN];
                for(int i=0; i<MAXN; i++) vis[i] = 0;
                BigInteger L, R;
                L = in.nextBigInteger();
                R = in.nextBigInteger();
                int dl = 0, dr = 0;
                dl = Low(L)+1;
                dr = Low(R)-1;
                long ans = fac[dr]-fac[dl-1];
                ans = (ans%MOD+MOD)%MOD;
                if(dl > dr) ans = 0; 
                dl--; dr++;
                
                long ans2=0;
                BigInteger tmp = (BigInteger.valueOf(dl)).pow(dl-1);
                for(int i=dl; i>=1; i--){
                    int top = L.divide(tmp).intValue();
                    if(i==dl && top==0) { ans2 = (ans2+fac[dl]-fac[dl-1])%MOD; break; }
                    int cnt=0;
                    for(int o=top+1;o<dl;o++) if(vis[o]==0) cnt++;
                    ans2 = (ans2+(cnt)*fac[i-1]%MOD)%MOD;
                    if(vis[top] == 1) break;
                    L = L.mod(tmp);
                    if(i==1 && vis[top]==0) ans2=ans2+1;
                    vis[top] = 1;
                    tmp = tmp.divide(BigInteger.valueOf(dl));
                }
                
                long ans1 = 0;
                for(int i=0; i<MAXN; i++) vis[i] = 0;
                tmp = (BigInteger.valueOf(dr)).pow(dr-1);
                for(int i=dr; i>=1; i--)
                {
                    int top = R.divide(tmp).intValue();
                    if(i==dr && top==0) {ans1 = (ans1+fac[dr]-fac[dr-1]%MOD); break;}
                    int cnt=0;
                    for(int o=top+1;o<dr;o++) if(vis[o]==0) cnt++;
                    ans1 = (ans1+(cnt)*fac[i-1]%MOD)%MOD;
                    if(vis[top] == 1) break;
                    R = R.mod(tmp);
                    vis[top] = 1;
                    tmp = tmp.divide(BigInteger.valueOf(dr));
                }
    //            System.out.println("ans1 -> "+ans1);
    //            System.out.println("ans2 -> "+ans2);
    //            System.out.println("ans  -> "+ans);
    //            System.out.println("DL -> "+dl);
    //            System.out.println("DR -> "+dr);
                ans = ans+ans2 + fac[dr]-fac[dr-1]-ans1;
                if(dl==dr) ans=ans2-ans1;
                ans = (ans%MOD+MOD)%MOD;
                System.out.println(ans);
            }
        }
    }
     
  • 相关阅读:
    RTThread | 启动下一代RTOS演化
    开发者应该开始学习C++吗?
    用googleperftool分析程序的内存/CPU使用
    看书看累了,可以换看技术视频也是一种学习的方式
    分享:nginx virtuanenv django1.4 应用简单部署
    分享:不同编程语言之间转换的项目矩阵
    【EDUPEPN8508GS黄金版】EDUP EPN8508GS黄金版 迷你USB无线网卡【行情 报价 价格 评测】
    分享:20 本优秀的 Python 电子书
    说说设计模式~工厂方法模式(Factory Method)
    说说设计模式~简单工厂模式(Factory)
  • 原文地址:https://www.cnblogs.com/chen9510/p/7308994.html
Copyright © 2011-2022 走看看