[kuangbin带你飞]专题十四 数论基础——A 题解

A - Bi-shoe and Phi-shoeTime Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %lluSubmit Status Practice LightOJ1370 uDebugDescriptionBamboo Pole-vault is a massively popular sport in Xzhiland. And Master Phi-shoe is a very popular coach for his success. He needs some bamboos for his students, so he asked his assistant Bi-Shoe to go to the market and buy them. Plenty of Bamboos of allpossible integer lengths (yes!) are available in the market. According to Xzhila tradition,Score of a bamboo = Φ (bamboo's length)(Xzhilans are really fond of number theory). For your information, Φ (n) = numbers less than n which are relatively prime (having no common divisor other than 1) to n. So, score of a bamboo of length 9 is6 as 1, 2, 4, 5, 7, 8 are relatively prime to 9.The assistant Bi-shoe has to buy one bamboo for each student. As a twist, each pole-vault student of Phi-shoe has a lucky number. Bi-shoe wants to buy bamboos such that each of them gets a bamboo with a score greater than or equal to his/her lucky number.Bi-shoe wants to minimize the total amount of money spent for buying the bamboos. One unit of bamboo costs 1 Xukha. Help him.InputInput starts with an integer T (≤ 100), denoting the number of test cases.Each case starts with a line containing an integer n (1 ≤ n ≤ 10000) denoting the number of students of Phi-shoe. The next line contains n space separated integers denoting the lucky numbers for the students. Each luckynumber will lie in the range [1, 106].OutputFor each case, print the case number and the minimum possible money spent for buying the bamboos. See the samples for details.Sample Input351 2 3 4 5610 11 12 13 14 1521 1Sample OutputCase 1: 22 XukhaCase 2: 88 XukhaCase 3: 4 Xukha题中明显要求欧拉函数，我就先把欧拉函数打了一张表出来然后在表中找比输入的d小的x的欧拉函数，提交后就无限超时。。。。后来写了一个式子，假设所求的数为k，数n的欧拉函数大等于k并且n最小一定满足一个式子：phi（n）＝（n＊（p1-1）(p2-1)*....*(pn-1)）/p1*p2*....pn＝k其中p为n的质因数那么一定有的是，k<＝n所以找的时候下限从k开始找就行了，唯独k==1的情况要单独讨论下面贴代码：#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cstdlib>using namespace std;const long long maxn=1000005;long long save[maxn];void init(){long long i,j;for(i=1;i<maxn;i++){save[i]=i;}for(i=2;i<maxn;i++){if(save[i]==i){for(j=i;j<maxn;j+=i){save[j]=save[j]*(i-1)/i;}}}}int main(){long long i,j,k,t,n,cnt=1;init();scanf("%lld",&t);while(t--){scanf("%lld",&n);long long result=0;for(i=1;i<=n;i++){scanf("%lld",&k);if(k==1){result=result+2;continue;}for(j=k;j<=maxn;j++){if(save[j]>=k){result += j;break;}}}printf("Case %lld: %lld Xukha\n",cnt,result);cnt++;}return 0;}