A - Bi-shoe and Phi-shoe
Crawling in process... Crawling failed 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
题意：求n个后面的每一个数的第一个素数并求和。
比较笨，先打表把素数标出来再取出来，再把每一位数后面的素数打出来。就这么胡来…
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <cmath>
5
6 using namespace std;
7
8 typedef long long LL;
9 const int maxn = 1000010;
10 bool isprime[maxn];
11 LL p[maxn];
12 LL m[maxn];
13
14 void printlist() {
15 memset(isprime, true, sizeof(isprime));
16 isprime[0] = isprime[1] = false;
17 int pedge = int(sqrt(maxn));
18 for(int i = 2; i <= pedge; i++) {
19 if(isprime[i]) {
20 int o = maxn / i;
21 for(int j = 2; j <= o; j++) {
22 isprime[i*j] = false;
23 }
24 }
25 }
26 }
27
28 void init() {
29 printlist();
30 int k = 0;
31 for(int i = 0; i < maxn; i++) {
32 if(isprime[i]) {
33 p[k++] = i;
34 }
35 }
36 k = 0;
37 for(int i = 0; i < maxn; i++) {
38 if(i < p[k]){
39 m[i] = p[k];
40 continue;
41 }
42 m[i] = p[++k];
43 }
44 }
45
46 int main() {
47 int T, n;
48 init();
49 scanf("%d", &T);
50 for(int ii = 1; ii <= T; ii++) {
51 scanf("%d", &n);
52 LL ans = 0;
53 int tmp;
54 while(n--) {
55 scanf("%d", &tmp);
56 ans += m[tmp];
57 }
58 printf("Case %d: %lld Xukha\n", ii, ans);
59 }
60 }
Hint