A-Bi-shoe and Phi-shoe
Time Limit: -Ms
Memory Limit:32768Kb
64bit IO Format:%lld &%lluSubmit Status
Description
Bamboo Pole-vault is a massively popular sport in Xzhiland. And Master Phi-shoe is a very popular coaches 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!) is available in the market. According to Xzhila tradition,
Score of a bamboo = Φ (bamboo ' s length)
(Xzhilans is really fond of number theory). For your information, Φ (n) = numbers less than n which is relatively prime (having no common divisor o Ther than 1) to N. So, score of a bamboo of length 9 is 6 as 1, 2, 4, 5, 7, 8 were relatively prime to 9.
The assistant Bi-shoe have to buy one bamboo for each student. As a twist, each pole-vault student of Phi-shoe have a lucky number. Bi-shoe wants to buy bamboos such, each of them gets a bamboo with a score greater than or equal to his/her lucky numb Er. 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 would 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
#include <stdio.h> #include <iostream> #include <string.h> #include <algorithm> #include < math.h> #include <ctype.h> #include <time.h> #include <queue> #include <iterator>using namespace Std;const int maxn = 1000100;int N, m, T;int PRIME[MAXN], PH[MAXN], P[MAXN], q[maxn];void solve (int n) {memset (pr Ime,0,sizeof (prime)); Memset (P,0,sizeof (P)); memset (ph,0,sizeof (ph)); int t = 0;for (int i = 2; I <= n; i++) {if (p[i] = = 0 ) Prime[++t] = i;for (int j = i * 2; j <= N; j+=i) {p[j] = 1;}} t = 1;for (int i = 1; I <= n; i++) {while (I >= prime[t]) {t++;} if (I < prime[t]) ph[i] = Prime[t];}} /*int phi[maxn];void Phi (int n) {for (int i = 0; I <= N; i++) phi[i] = 0;phi[1] = 1;for (int i = 2; I <= n; i++) {if (! Phi[i]) {for (int j = i; J <= N; j + = i) {if (!phi[j]) phi[j] = j;phi[j] = phi[j]/i* (i-1);}}} */int Main () {int cases = 1;solve (1001000), scanf ("%d", &t), while (t--) {long-long ans = 0;scanf ("%d", &n); for (int i = 0; I < n; i++) SCANF ("%d", &q[i]); sort (q,q+n); for (int i = 0; i < n; i++) ans + ph[q[i]];p rintf ("Case%d:%lld xukha\n", Cases++,ans); }return 0;}
BNU 13288 Bi-shoe and Phi-shoe "prime filter"