ZOJ - 2358 Sum of Factorials

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本文深入探讨了算法和数据结构在计算机科学领域的核心应用，涵盖了从基础的排序算法到复杂的深度学习模型，以及在大数据处理、人工智能、区块链等多个前沿技术领域的实践案例。文章不仅解析了各种算法的工作原理，还提供了实际应用场景的分析，旨在帮助读者理解这些技术如何解决实际问题。

Sum of Factorials

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

Submit Status

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 SUM{xi!} (1 <= i <= t, 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. 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

Source

Asia 2003, Guangzhou (Mainland China)

分析:

简单题。可以证明，a[i]!>=a[0]+a[1]+......+a[i-1](等号当且仅当i=1成立)。所以只要将n每次减去最大的a[i]!(如果n>=a[i]!的话)。这样大大简化了程序，避免了dfs搜索。

看来动一下脑筋有时候可以省很大力气。

ac代码：

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

int main()
{
    int a[11];
    int i,n;
    a[0]=1;
    for(i=1;i<11;i++)
    a[i]=a[i-1]*i;
    while(scanf("%d",&n)&&n>=0)
    {
        if(n==0)
        {
            printf("NO\n");
            continue;
        }
        for(i=10;i>=0&&n>0;i--)
        {
            if(n>=a[i])
            n-=a[i];
        }
        if(n==0)
        printf("YES\n");
        else
        printf("NO\n");
    }
    return 0;
}