zoukankan      html  css  js  c++  java
  • HDOJ 3944 DP?

    尽量沿着边走距离最短。化减后 C(n+1,k)+ n - k,

    预处理阶乘,Lucas定理组合数取模


    DP?

    Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 128000/128000 K (Java/Others)
    Total Submission(s): 1899    Accepted Submission(s): 633


    Problem Description

    Figure 1 shows the Yang Hui Triangle. We number the row from top to bottom 0,1,2,…and the column from left to right 0,1,2,….If using C(n,k) represents the number of row n, column k. The Yang Hui Triangle has a regular pattern as follows.
    C(n,0)=C(n,n)=1 (n ≥ 0) 
    C(n,k)=C(n-1,k-1)+C(n-1,k) (0<k<n)
    Write a program that calculates the minimum sum of numbers passed on a route that starts at the top and ends at row n, column k. Each step can go either straight down or diagonally down to the right like figure 2.
    As the answer may be very large, you only need to output the answer mod p which is a prime.
     

    Input
    Input to the problem will consists of series of up to 100000 data sets. For each data there is a line contains three integers n, k(0<=k<=n<10^9) p(p<10^4 and p is a prime) . Input is terminated by end-of-file.
     

    Output
    For every test case, you should output "Case #C: " first, where C indicates the case number and starts at 1.Then output the minimum sum mod p.
     

    Sample Input
    1 1 2 4 2 7
     

    Sample Output
    Case #1: 0 Case #2: 5
     

    Author
    phyxnj@UESTC
     

    Source
     


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    
    typedef long long int LL;
    
    LL n,k,p;
    
    LL fact[1300][11000];
    
    LL QuickPow(LL x,LL t,LL m)
    {
    	if(t==0) return 1LL;
    	LL e=x,ret=1LL;
    	while(t)
    	{
    		if(t&1LL) ret=(ret*e)%m;
    		e=(e*e)%m;
    		t>>=1LL;
    	}
    	return ret%m;
    }
    
    int prime[2000],pr;
    bool vis[10100];
    
    void get_prime()
    {
    	for(int i=2;i<10100;i++)
    	{
    		if(vis[i]==false)	
    			prime[pr++]=i;
    		for(int j=2*i;j<10100;j+=i)
    			vis[j]=true;
    	}
    }
    
    void get_fact()
    {
    	for(int i=0;i<1240;i++)
    	{
    		fact[i][0]=1LL;
    		for(int j=1;j<=prime[i]+10;j++)	
    		{
    			fact[i][j]=(fact[i][j-1]*j)%prime[i];	
    		}
    	}
    }
    
    LL Lucas(LL n,LL m,LL p)
    {
    	LL ret=1LL;
    	int id=lower_bound(prime,prime+pr,p)-prime;
    	while(n&&m)
    	{
    		LL a=n%p,b=m%p;
    		if(a<b) return 0;
    		ret=(ret*fact[id][a]*QuickPow((fact[id][b]*fact[id][a-b])%p,p-2,p)%p)%p;
    		n/=p; m/=p;
    	}
    	return ret%p;
    }
    
    int main()
    {
    	get_prime();
    	get_fact();
    	int cas=1;
    	while(scanf("%I64d%I64d%I64d",&n,&k,&p)!=EOF)
    	{
    		if(k>n/2) k=n-k;
    		LL ans=(Lucas(n+1,k,p)+n-k)%p;
    		printf("Case #%d: %I64d
    ",cas++,ans);
    	}
    	return 0;
    }




  • 相关阅读:
    Android stadio 插件推荐--ok gradle
    算法:枚举法---kotlin
    Kotlin 二分法算法游戏--猜价格
    android onCreate的两个方法
    Spring 中的scope
    Intellij IDEA 2017 debug断点调试技巧与总结详解篇
    深入浅出ConcurrentHashMap1.8
    ConcurrentHashMap JDK1.8
    synchronized修饰普通方法和静态方法
    Java多线程系列--CopyOnWriteArraySet
  • 原文地址:https://www.cnblogs.com/yfceshi/p/6780946.html
Copyright © 2011-2022 走看看