| 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 somebamboos for his students, so he asked his assistant Bi-Shoe to go to the marketand buy them. Plenty of Bamboos of all possible integer lengths (yes!) areavailable in the market. According to Xzhila tradition,
Score of a bamboo = Φ (bamboo's length)
(Xzhilans are really fond of number theory). For yourinformation, Φ (n) = numbers less thann which arerelatively prime (having no common divisor other than 1) ton. So, scoreof 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 eachstudent. 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 scoregreater than or equal to his/her lucky number. Bi-shoe wants to minimize thetotal amount of money spent for buying the bamboos. One unit of bamboo costs 1Xukha. 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 linecontainsn space separated integers denoting the lucky numbers for thestudents. Each lucky number will lie in the range[1, 106].
Output
For each case, print the case number and the minimumpossible 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 |
題意是知道一串數字。然後規定一個規則，score 爲 length以內與length互質的數的個數，每個數字依次分析，對於每個數字，求不小於這個數的score對應的length，然後加起來。
這個題可以轉化一下題意，我們知道對於一個length爲質數的數，他的score一定等於length - 1，然後對於任意一個合數，他的score一定小於它前面的那個質數的score，所以題意就轉化爲求解比當前數字大的最小的質數的大小，將所有的累計，得到最後的答案。
歐拉篩法解決。
代碼如下：
/*************************************************************************
> File Name: Bi-shoe_and_Phi-shoe.cpp
> Author: Zhanghaoran
> Mail: [email protected]
> Created Time: Tue 08 Dec 2015 12:38:03 PM CST
************************************************************************/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int Max = 1000100;
int Prim[Max];
int Isprim[Max];
void euler(){
int ans = 0;
Isprim[0] = Isprim[1] = 1;
for(int i = 2; i <= Max - 1; i ++){
if(!Isprim[i])
Prim[ans ++] = i;
for(int j = 0; j < ans && Prim[j] * i <= Max - 1; j ++){
Isprim[Prim[j] * i] = 1;
if(i % Prim[j] == 0)
break;
}
}
}
int T;
int N;
int main(void){
cin >> T;
long sum = 0;
euler();
int cas = 0;
while(T --){
cin >> N;
cas ++;
int temp;
for(int i = 0; i < N; i ++){
cin >> temp;
for(int j = temp + 1; j <= Max - 1; j ++){
if(!Isprim[j]){
sum += j;
break;
}
}
}
printf("Case %d: %ld Xukha\n", cas, sum);
sum = 0;
}
}