By 2312 there were n Large Hadron Colliders in the inhabited part of the universe. Each of them corresponded to a single natural number from 1 to n. However, scientists did not know what activating several colliders simultaneously could cause, so the colliders were deactivated.
In 2312 there was a startling discovery: a collider‘s activity is safe if and only if all numbers of activated colliders are pairwise relatively prime to each other (two numbers are relatively prime if their greatest common divisor equals 1)! If two colliders with relatively nonprime numbers are activated, it will cause a global collapse.
Upon learning this, physicists rushed to turn the colliders on and off and carry out all sorts of experiments. To make sure than the scientists‘ quickness doesn‘t end with big trouble, the Large Hadron Colliders‘ Large Remote Control was created. You are commissioned to write the software for the remote (well, you do not expect anybody to operate it manually, do you?).
Initially, all colliders are deactivated. Your program receives multiple requests of the form "activate/deactivate the i-th collider". The program should handle requests in the order of receiving them. The program should print the processed results in the format described below.
To the request of "+ i" (that is, to activate the i-th collider), the program should print exactly one of the following responses:
The request of "- i" (that is, to deactivate the i-th collider), should receive one of the following responses from the program:
You don‘t need to print quotes in the output of the responses to the requests.
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of colliders and the number of requests, correspondingly.
Next m lines contain numbers of requests, one per line, in the form of either "+ i" (without the quotes) — activate the i-th collider, or "- i" (without the quotes) — deactivate the i-th collider (1 ≤ i ≤ n).
Print m lines — the results of executing requests in the above given format. The requests should be processed in the order, in which they are given in the input. Don‘t forget that the responses to the requests should be printed without quotes.
10 10
+ 6
+ 10
+ 5
- 10
- 5
- 6
+ 10
+ 3
+ 6
+ 3
Success
Conflict with 6
Success
Already off
Success
Success
Success
Success
Conflict with 10
Already on
Note that in the sample the colliders don‘t turn on after the second and ninth requests. The ninth request could also receive response "Conflict with 3".
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long 10 #define INF 0x3f3f3f 11 using namespace std; 12 const int maxn = 100010; 13 int prime[maxn],index[maxn],tot = 0; 14 bool npm[maxn] = {true,true}; 15 bool vis[maxn]; 16 vector<int>pm[maxn]; 17 void getprime() { 18 int i,j,temp; 19 for(i = 2; i < maxn; i++) { 20 if(!npm[i]) prime[tot++] = i; 21 for(j = 0; j < tot && (temp = i*prime[j]) < maxn; j++) { 22 npm[temp] = true; 23 if(i%prime[j] == 0) break; 24 } 25 } 26 for(i = 2; i < maxn; i++) { 27 if(!npm[i]) pm[i].push_back(i); 28 else { 29 int t = sqrt(i),k = i; 30 for(j = 0; j < tot && prime[j] <= t; j++) { 31 if(k%prime[j] == 0) { 32 pm[i].push_back(prime[j]); 33 while(k && k%prime[j] == 0) k /= prime[j]; 34 } 35 } 36 if(k > 1) pm[i].push_back(k); 37 } 38 } 39 } 40 void del(int u){ 41 for(int i = 0; i < pm[u].size(); i++) 42 index[pm[u][i]] = -1; 43 } 44 bool calc(int x,int &op) { 45 int i,j; 46 bool flag = true; 47 for(i = 0; i < pm[x].size(); i++) { 48 if(index[pm[x][i]] > 0) { 49 flag = false; 50 op = index[pm[x][i]]; 51 break; 52 } 53 } 54 if(flag) { 55 for(i = 0; i < pm[x].size(); i++) 56 index[pm[x][i]] = x; 57 } 58 return flag; 59 } 60 int main() { 61 int n,m,id,i; 62 char s[4]; 63 getprime(); 64 while(~scanf("%d %d",&n,&m)){ 65 memset(vis,false,sizeof(vis)); 66 memset(index,-1,sizeof(index)); 67 for(i = 0; i < m; i++){ 68 scanf("%s %d",s,&id); 69 if(id == 1){ 70 if(s[0] == ‘+‘){ 71 if(vis[id]) puts("Already on"); 72 else {vis[id] = true;puts("Success");} 73 }else{ 74 if(vis[id]) {vis[id] = false;puts("Success");} 75 else puts("Already off"); 76 } 77 }else{ 78 if(s[0] == ‘+‘){ 79 if(vis[id]) puts("Already on"); 80 else{ 81 int conflic; 82 if(calc(id,conflic)){ 83 vis[id] = true; 84 puts("Success"); 85 }else printf("Conflict with %d\n",conflic); 86 } 87 }else{ 88 if(vis[id]) {del(id);vis[id] = false;puts("Success");} 89 else puts("Already off"); 90 91 } 92 } 93 } 94 } 95 return 0; 96 }
xtu summer individual 2 D - Colliders,布布扣,bubuko.com
xtu summer individual 2 D - Colliders
原文:http://www.cnblogs.com/crackpotisback/p/3885729.html