LightOJ 1370-Bi-shoe and Phi-shoe（欧拉函数）

A - Bi-shoe and Phi-shoe
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Description
Bamboo 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 all possible 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 is 6 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.

Input
Input 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 lucky number will lie in the range [1, 106].

Output
For each case, print the case number and the minimum possible money spent for buying the bamboos. See the samples for details.

Sample Input
3
5
1 2 3 4 5
6
10 11 12 13 14 15
2
1 1
Sample Output
Case 1: 22 Xukha
Case 2: 88 Xukha
Case 3: 4 Xukha

题目大意：
首先你要明白一个数的价值的定义。给你一个数n，假设小于它的数有x个与它互质，我们称它的价值是x。
给你一个价值，让你找对应的一个数，这个数的价值大于等于题目所给的价值，且必须最小，题目一共给你n个价值，你要找n个数，并加和就好了。
最朴素的做法是根据定义找每个数的价值，然后打成一张表，然后一个个去比对，但是那样会超时。
所以可以用欧拉函数，它将直接为你打印一张每个数的价值表，当我们知道一个数x的价值时，直接从表中的x位置去搜寻，直到找到一个位置，那个位置的值大于即可。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int euler[1200000];void getEuler();int main(){    //freopen("in.txt","r",stdin);    getEuler();    euler[1]=0;    int n;    scanf("%d",&n);    long long int sum=0;    for(int w=1; w<=n; w++)    {        int b;        scanf("%d",&b);        sum=0;        for(int i=1; i<=b; i++)        {            int q;            scanf("%d",&q);            int j=q-2;            if(j<=0) j=1;            for(;j<=1100000;j++)            {                //cout<<j<<"  "<<euler[j]<<endl;                if(euler[j]>=q)                {                    sum+=j;                    break;                }            }        }        printf("Case %d: %lld Xukha\n",w,sum);    }    return 0;}void getEuler(){    memset(euler,0,sizeof(euler));    euler[1]=1;    for(int i=2; i<=1100000; i++)        if(!euler[i])            for(int j=i; j<=1100000; j+=i)            {                if(!euler[j])                    euler[j]=j;                euler[j]=euler[j]/i*(i-1);            }}