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  • 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;
    }


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