Light oj 1307 - Bi-shoe and Phi-shoe【欧拉函数】
2016-09-13 17:30
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1370 - Bi-shoe and Phi-shoe
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.
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].
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。
PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
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\n", ++Kcase, res); } return 0; }
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