首页 > 其他 > 详细

HDOJ 4349 DP?

时间:2014-08-11 02:58:51      阅读:321      评论:0      收藏:0      [点我收藏+]

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
bubuko.com,布布扣

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\n",cas++,ans);
	}
	return 0;
}




HDOJ 4349 DP?,布布扣,bubuko.com

HDOJ 4349 DP?

原文:http://blog.csdn.net/ck_boss/article/details/38479839

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!