Light oj 1307 - Bi-shoe and Phi-shoe【欧拉函数】

1370 - Bi-shoe and Phi-shoe

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Time Limit: 2 second(s)Memory Limit: 32 MB

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

Output for Sample Input

3

5

1 2 3 4 5

6

10 11 12 13 14 15

2

1 1

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

 

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题意：

每次给你n个数，这些数看做欧拉函数的值。 每一个值会对应多个自变量，求出每一个自变量的最小值，并对他们求值。每个自变量对应的函数值大于等于给出的数。

思路：

素数p所对应的欧拉函数值为p-1，此时对于p-1来说，所对应的自变量最小的就是p。所以对于任意的一个正整数q，q所对的自变量的最小值必定为q之后的第一个素数。

然后进行素数打表判断。

#include <map>#include <set>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>#include <stack>#include <cmath>#include <string>#include <vector>#include <cstdlib>//#include <bits/stdc++.h>//#define LOACL#define space " "using namespace std;typedef long long Long;//typedef __int64 Int;typedef pair<int, int> paii;const int INF = 0x3f3f3f3f;const double ESP = 1e-5;const double Pi = acos(-1.0);const int MOD = 1e9 + 5;const int MAXN = 1e6 + 5;bool nprime[MAXN];void init() {memset(nprime, false, sizeof(nprime));nprime[0] = nprime[1] =  true;for (int i = 2; i < MAXN; i++) {if (!nprime[i]) {for (int j = i*2; j < MAXN; j += i) nprime[j] = true;}}}int main() {int t, a, n;int Kcase = 0;scanf("%d", &t);while (t--) {init(); Long res = 0;scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &a);int cnt = a + 1;while (true) {if (!nprime[cnt]) {res += cnt; break;}cnt++;}}printf("Case %d: %lld Xukha\n", ++Kcase, res);}return 0;}