UVALive - 6575 Odd and Even Zeroes 数位dp+找规律

题目链接：

http://acm.hust.edu.cn/vjudge/problem/48419

Odd and Even Zeroes

Time Limit: 3000MS #### 问题描述 > In mathematics, the factorial of a positive integer number n is written as n! and is defined as follows: > n! = 1 × 2 × 3 × 4 × . . . × (n − 1) × n = > ∏n > i=1 > i > The value of 0! is considered as 1. n! grows very rapidly with the increase of n. Some values of n! > are: > 0! = 1 > 1! = 1 > 2! = 2 > 3! = 6 > 4! = 24 > 5! = 120 > 10! = 3628800 > 14! = 87178291200 > 18! = 6402373705728000 > 22! = 1124000727777607680000 > You can see that for some values of n, n! has odd number of trailing zeroes (eg 5!, 18!) and for some > values of n, n! has even number of trailing zeroes (eg 0!, 10!, 22!). Given the value of n, your job is to > find how many of the values 0!, 1!, 2!, 3!, . . . ,(n − 1)!, n! has even number of trailing zeroes. > #### 输入 > Input file contains at most 1000 lines of input. Each line contains an integer n (0 ≤ n ≤ 1018). Input > is terminated by a line containing a ‘-1’.

输出

For each line of input produce one line of output. This line contains an integer which denotes how
many of the numbers 0!, 1!, 2!, 3!, . . . , n!, contains even number of trailing zeroes.

样例

sample input

2
3
10
100
1000
2000
3000
10000
100000
200000
-1

sample output
3
4
6
61
525
1050
1551
5050
50250
100126

题意

求0!,1!,...,n!里面末尾有偶数个零的数的个数。

题解

将n按五进制展开，发现如果只有当偶数位权上的数的和为偶数时，n的末尾有偶数个0。所以将问题转换成统计小于n的偶数位权为偶数的数有多少个。
这个用数位dp可以解决。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
#define bug(x) cout<<#x<<" = "<<x<<endl;
using namespace std;

const int maxn = 66;
typedef long long  LL;

int arr[maxn],tot;

//dp[i][0]表示前i位中偶数位上的和为偶数的数的个数 
//dp[i][1]表示前i位中偶数位上的和为奇数的数的个数 
LL dp[maxn][2];
	LL dfs(int len, int type,bool ismax,bool iszer) {
	if (len == 0) {
	if(!type) return 1LL;
			else return 0LL; 
	}
	if (!ismax&&dp[len][type]>0) return dp[len][type];
	LL res = 0;
	int ed = ismax ? arr[len] : 4;

	for (int i = 0; i <= ed; i++) {
		if(len&1){
			res+=dfs(len-1,type,ismax&&i == ed,iszer&&i==0);
		} 
		else{
			if((i&1)) res+=dfs(len-1,type^1,ismax&&i == ed,iszer&&i==0);
			else res+=dfs(len-1,type,ismax&&i == ed,iszer&&i==0);
		}
	}
	return ismax ? res : dp[len][type] = res;
}

LL solve(LL x) {
	tot = 0;
	//五进制
	while (x) { arr[++tot] = x % 5; x /= 5; }
	return dfs(tot,0, true,true);
}

int main() {
	LL x;
	memset(dp,-1,sizeof(dp));
	while (scanf("%lld",&x)==1&&x!=-1) {
		printf("%lld\n", solve(x));
	}
	return 0;
}