Light oj 1307

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


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
", ++Kcase, res);
    }
    return 0;
}


原文地址:https://www.cnblogs.com/cniwoq/p/6770798.html