A prime number p≥2 is an integer which is evenly divisible by only two integers: 1 and p. A composite integer is one which is not prime. The fundamental theorem of arithmetic says that any integer x can be expressed uniquely as a set of prime factors – those prime numbers which, when multiplied together, give x. Consider the prime factorization of the following numbers:
10=2×5 16=2×2×2×2 231=3×7×11
Consider the following process, which we’ll call prime reduction. Given an input x:

if xx is prime, print xx and stop
factor xx into its prime factors p1,p2,…,pk
let x=p1+p2+?+pk
go back to step 1
Write a program that implements prime reduction.

Input
Input consists of a sequence of up to 2000020000 integers, one per line, in the range 22 to 109109. The number 44 will not be included in the sequence (try it to see why it’s excluded). Input ends with a line containing only the number 4.

Output
For each integer, print the value produced by prime reduction executed on that input, followed by the number of times the first line of the process executed.

Sample Input 1Sample Output 1

2 3 5 76 100 2001 4	2 1 3 1 5 1 23 2 5 5 5 6

大意就是：給你一個數(shù)x——1、如果x是素數(shù)，直接輸出x以及循環(huán)的步數(shù)。2、如果不是，那就x分解質因數(shù)，把所有質因數(shù)之和給x，步數(shù)+1，執(zhí)行第一步。

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//Asimple #include <bits/stdc++.h> using namespace std; typedef long long ll; ll n, m, s, res, ans, len, T, k, num;

bool is_pr(ll n) {for(int i=2; i*i<=n; i++) {if( n%i==0 ) return false;}return true; }ll solve(ll n){ll ans = 0;int i = 2;while( n>1 ) {if( n%i==0 ) {ans += i;n /= i;if( is_pr(n) ) {//這步是關鍵，不寫超時ans += n;break;} } else ++i;}return ans; } void input() {while( cin >> n) {res = 1;if( n == 4 ) break;while( !is_pr(n) ) {n = solve(n);res ++;}cout << n << " " << res << endl;} }int main(){input();return 0; }

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轉載于:https://www.cnblogs.com/Asimple/p/6785476.html

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