【2008IO】#ULR 2 A. Goshtasp, Vishtasp and Eidi

最新推荐文章于 2023-11-28 18:02:22 发布

原创最新推荐文章于 2023-11-28 18:02:22 发布 · 676 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Codeforces

CF-A题终结者专栏收录该内容

96 篇文章

订阅专栏

本博客介绍了一个关于识别富数的任务，通过输入正整数n来判断其是否为富数，并输出构成该数的质数或1的序列。提供了解决此问题的方法，包括输入验证、算法实现及输出格式说明，旨在帮助参与者理解如何解决此类数学问题并通过提供答案获得奖励。

Goshtasp was known to be a good programmer in his school. One day Vishtasp, Goshtasp's friend, asked him to solve this task:

Given a positive integer n, you should determine whether n is rich.

The positive integer x is rich, if there exists some set of distinct numbers a1, a2, ..., am such that $\text{[math]}$ . In addition: every ai should be either a prime number, or equal to 1.

Vishtasp said that he would share his Eidi 50 / 50 with Goshtasp, if he could solve the task. Eidi is money given to children for Noruz by their parents and/or relatives.

Goshtasp needs to solve this problem to get money, you need to solve it to get score!

Input

Input contains a single positive integer n (1 ≤ n ≤ 10000).

Output

If the number is not rich print 0. Otherwise print the numbers a1, ..., am. If several solutions exist print the lexicographically latestsolution. Answers are compared as sequences of numbers, not as strings.

For comparing two sequences a1, ..., am and b1, ..., bn we first find the first index i such that ai ≠ bi, if ai < bi then a is lexicographically earlier and if bi < ai then b is lexicographically earlier. If m ≠ n we add zeroes at the end of the smaller sequence (only for the moment of comparison) and then perform the comparison.

You do not need to minimize the number of elements in sequence (i.e. m). You just need to print the lexicographically latestsolution.

See samples to find out how to print the sequence.

Sample test(s)

input
11

output
11=11

input
545

output
541+3+1=545

又是一场Unknown Language Round……没办法 A题终结者的No.是按照Round编号来的（啊那个加入专栏的不是按Round号而是按照#号顺序的哦~）

强迫症表示，没什么学习意义，不过是也贴一下完成强迫症的心情：

n :=File standardInput readLine asNumber
p :=0
m :=n
while(n>0,for(x,n,1,-1,
q :=0
for(i,2,x-1,if(x%i==0,q :=1;break;))
if(q==0,
if(p==1,"+" print)
p :=1
n :=n-x
x print
break
)
))
"=" print
m println