POJ 1775 FZU 1106 LightOJ 1189 Sum of Factorials

POJ 1775  、 FZU 1106     Sum of Factorials 

题目：

Description

John von Neumann, b. Dec. 28, 1903, d. Feb. 8, 1957, was a Hungarian-American mathematician who made important contributions to the foundations of mathematics, logic, quantum physics,meteorology, science, computers, and game theory. He was noted for a phenomenal memory and the speed with which he absorbed ideas and solved problems. In 1925 he received a B.S. diploma in chemical engineering from Zurich Institute and in 1926 a Ph.D. in mathematics from the University of Budapest. His Ph.D. dissertation on set theory was an important contribution to the subject. At the age of 20, von Neumann proposed a new definition of ordinal numbers that was universally adopted. While still in his twenties, he made many contributions in both pure and applied mathematics that established him as a mathematician of unusual depth. His Mathematical Foundations of Quantum Mechanics (1932) built a solid framework for the new scientific discipline. During this time he also proved the mini-max theorem of GAME THEORY. He gradually expanded his work in game theory, and with coauthor Oskar Morgenstern he wrote Theory of Games and Economic Behavior (1944). 
There are some numbers which can be expressed by the sum of factorials. For example 9,9=1!+2!+3! Dr. von Neumann was very interested in such numbers. So, he gives you a number n, and wants you to tell him whether or not the number can be expressed by the sum of some factorials. 
Well, it's just a piece of cake. For a given n, you'll check if there are some xi, and let n equal to Σ  _1<=i<=tx  _i!. (t >=1 , xi >= 0, xi = xj iff. i = j). If the answer is yes, say "YES"; otherwise, print out "NO".

Input

You will get several non-negative integer n (n <= 1,000,000) from input file. Each one is in a line by itself. 
The input is terminated by a line with a negative integer.

Output

For each n, you should print exactly one word ("YES" or "NO") in a single line. No extra spaces are allowed.

Sample Input

9
-1

Sample Output

YES

这个题目是考1个性质：

当n>1时，0！+1！+2！+......+n!<(n+1)!，当n=1时，0！+1！=2！,当n=0时，0！=1！

任取一个递增的函数f(x)满足下述条件：当x为整数时，f(x)=x!

用g表示f的反函数，那么这个算法的时间复杂度为θ(g(n))

比如说，本题n约为10！，那么大约有10次运算

（我这个描述其实是不准确的，因为有常数的问题。

准确的说法应该是，这个算法有大约g(n)次运算，常数为1）

代码：

#include<iostream>
using namespace std;

int fac[10] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 };

int main()
{
	int n;
	while (cin >> n)
	{
		if (n < 0)break;
		if (n == 0)cout << "NO";
		else
		{
			for (int i = 9; i >= 0; i--)if (n>=fac[i])n -= fac[i];
			if (n == 0)cout << "YES";
			else cout << "NO";
		}
		cout << endl;
	}
	return 0;
}

LightOJ 1189   Sum of Factorials 

题目：

Description

Given an integer n, you have to find whether it can be expressed as summation of factorials. For given n, you have to report a solution such that

n = x₁! + x₂! + ... + x_n! (x_i < x_j for all i < j)

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10¹⁸).

Output

For each case, print the case number and the solution in summation of factorial form. If there is no solution then print 'impossible'. There can be multiple solutions, any valid one will do. See the samples for exact formatting.

Sample Input

4

7

7

9

11

Sample Output

Case 1: 1!+3!

Case 2: 0!+3!

Case 3: 1!+2!+3!

Case 4: impossible

题目差不多，代码稍微改下就可以了。

代码：

#include<iostream>
#include<stack>
using namespace std;

long long fac[20] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800,87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 121645100408832000 };
stack<int>s;

int main()
{
	int t;
	cin >> t;
	long long n;
	for (int cas = 1; cas <= t; cas++)
	{		
		cin >> n;
		cout << "Case " << cas << ": ";
		while (!s.empty())s.pop();
		for (int i = 19; i >= 0; i--)if (n >= fac[i])
		{
			n -= fac[i];
			s.push(i);
		}
		if (n == 0)
		{
			while (!s.empty())
			{
				n = s.top();
				s.pop();
				cout << n << '!';
				if (!s.empty())cout << '+';
			}
		}
		else cout << "impossible";
		cout << endl;		
	}	
	return 0;
}