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| 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
Source
/** 题意：f(n) 表示 小于等于 n 的数中素数的个数；给出一串数 比如x 求f(x) >= x 的最小和 做法：欧拉函数 **/ #include <iostream> #include<cmath> #include<string.h> #include<stdio.h> #include<algorithm> #include<stack> #define maxn 1000000 + 10 int mindiv[maxn],phi[maxn],sum[maxn]; int mmap[maxn]; int mmpp[maxn]; using namespace std; void solve() { for(int i=1; i<maxn; i++) { mindiv[i] = i; } for(int i=2; i*i<maxn; i++) { if(mindiv[i] == i) { for(int j=i*i; j<maxn; j+=i) { mindiv[j] = i; } } } phi[1] = 1; for(int i=2; i<maxn; i++) { phi[i] = phi[i/mindiv[i]]; if((i/mindiv[i])%mindiv[i] == 0) { phi[i] *=mindiv[i]; } else { phi[i] *= mindiv[i] -1; } } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif // ONLINE_JUDGE int T; scanf("%d",&T); int n; solve(); int Case = 1; while(T--) { scanf("%d",&n); int res = 0; long long sum = 0; for(int i=0; i<n; i++) { scanf("%d",&mmpp[i]); } sort(mmpp,mmpp+n); int tt= 0,Index = 0; int j = 2; phi[1] = 1; for(int i=0; i<n; ) { if(phi[j] >= mmpp[i]) { sum += j; i++; } else j++; } printf("Case %d: %lld Xukha\n",Case++,sum); } return 0; }
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原文地址：http://www.cnblogs.com/chenyang920/p/4445555.html
