On
an alien planet, every extraterrestrial is born with a number. If the
sum of two numbers is a prime number, then two extraterrestrials can be
friends. But every extraterrestrial can only has at most one friend. You
are given all number of the extraterrestrials, please determining the
maximum number of friend pair.
There are several test cases.
Each test start with positive integers N(1 ≤ N ≤ 100), which means there are N extraterrestrials on the alien planet.
The following N lines, each line contains a positive integer pi ( 2 ≤ pi
≤10^18),indicate the i-th extraterrestrial is born with pi number.
The input will finish with the end of file.
For each the case, your program will output maximum number of friend pair.
1 #include <cstdio>
2 #include <cstring>
3 #include <cmath>
4 #include <algorithm>
5 #include <string>
6 #include <vector>
7 #include <set>
8 #include <map>
9 #include <stack>
10 #include <queue>
11 #include <sstream>
12 #include <iomanip>
13 #include <ctime>
14 using namespace std;
15 typedef long long LL;
16 const int INF=0x4fffffff;
17 const int EXP=1e-5;
18 const int MS=105;
19
20 int vis[MS];
21 int edge[MS][MS];
22
23 int lin[MS];
24 LL a[MS];
25 int n;
26
27
28 //
29 LL MIN;
30 LL mult_mod(LL a,LL b,LL n)
31 {
32 LL s=0;
33 while(b)
34 {
35 if(b&1) s=(s+a)%n;
36 a=(a+a)%n;
37 b>>=1;
38 }
39 return s;
40 }
41
42 LL pow_mod(LL a,LL b,LL n)
43 {
44 LL s=1;
45 while(b)
46 {
47 if(b&1) s=mult_mod(s,a,n);
48 a=mult_mod(a,a,n);
49 b>>=1;
50 }
51 return s;
52 }
53
54 bool Prime(LL n)
55 {
56 LL u=n-1,pre,x;
57 int i,j,k=0;
58 if(n==2||n==3||n==5||n==7||n==11) return 1;
59 if(n==1||(!(n%2))||(!(n%3))||(!(n%5))||(!(n%7))||(!(n%11))) return 0;
60 for(;!(u&1);k++,u>>=1);
61 srand((LL)time(0));
62 for(i=0;i<5;i++)
63 {
64 x=rand()%(n-2)+2;
65 x=pow_mod(x,u,n);
66 pre=x;
67 for(j=0;j<k;j++)
68 {
69 x=mult_mod(x,x,n);
70 if(x==1&&pre!=1&&pre!=(n-1))
71 return 0;
72 pre=x;
73 }
74 if(x!=1) return false;
75 }
76 return true;
77 }
78
79 bool dfs(int x) // dfs增广路
80 {
81 for(int i=1;i<=n;i++)
82 {
83 if(edge[x][i]&&!vis[i])
84 {
85 vis[i]=1;
86 if(lin[i]==-1|| dfs(lin[i])) // i没有匹配 或匹配了可以得到增广路
87 {
88 lin[i]=x;
89 return true;
90 }
91 }
92 }
93 return false;
94 }
95
96 int main()
97 {
98 while(scanf("%d",&n)!=EOF)
99 {
100 memset(edge,0,sizeof(edge));
101 for(int i=1;i<=n;i++)
102 scanf("%lld",&a[i]);
103 for(int i=1;i<=n;i++)
104 for(int j=i+1;j<=n;j++)
105 if(Prime(a[i]+a[j]))
106 edge[i][j]=edge[j][i]=1;
107 int ans=0;
108 memset(lin,-1,sizeof(lin));
109 for(int i=1;i<=n;i++)
110 {
111 memset(vis,0,sizeof(vis));
112 if(dfs(i))
113 ans++;
114 }
115 printf("%d\n",ans/2);
116 }
117 return 0;
118 }