1552: Friends
Time Limit: 3 Sec Memory Limit: 256 MBSubmit: 187 Solved: 43
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Description
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.
Input
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.
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.
Output
For each the case,your program will output maximum number of friend pair.
Sample Input
3
2
2
3
4
2
5
3
8
Sample Output
1 2
HINT
当遇到匹配问题时，而且只能用一次，这时应想一下二分最大匹配。
#include<stdio.h> #include<math.h> #include <algorithm> #include<string.h> #include <time.h> using namespace std; #define ll long long //-----大素数的判断------- ll mult_mod(ll a,ll b,ll n) { ll s=0; while(b) { if(b&1) s=(s+a)%n; a=(a+a)%n; b>>=1; } return s; } ll pow_mod(ll a,ll n) { ll s=1; while(b) { if(b&1) s=mult_mod(s,a,n); a=mult_mod(a,n); b>>=1; } return s; } int Prime(ll n) { ll u=n-1,pre,x; int i,j,k=0; if(n==2||n==3||n==5||n==7||n==11) return 1; if(n==1||(!(n%2))||(!(n%3))||(!(n%5))||(!(n%7))||(!(n%11))) return 0; for(;!(u&1);k++,u>>=1); srand((ll)time(0)); for(i=0;i<5;i++) { x=rand()%(n-2)+2; x=pow_mod(x,u,n); pre=x; for(j=0;j<k;j++) { x=mult_mod(x,x,n); if(x==1&&pre!=1&&pre!=(n-1)) return 0; pre=x; } if(x!=1) return 0; } return 1; } int n,ok[105][105],mark[105],vist[105]; int DFS(int x) { for(int i=1;i<=n;i++) if(ok[x][i]&&vist[i]==0) { vist[i]=1; if(mark[i]==0||DFS(mark[i])) { mark[i]=x; return 1; } } return 0; } int main() { ll a[105]; while(scanf("%d",&n)>0) { for(int i=1;i<=n;i++) { scanf("%lld",&a[i]); } memset(ok,sizeof(ok)); for(int i=1;i<n;i++) for(int j=i+1;j<=n;j++) if(Prime(a[i]+a[j])) ok[i][j]=ok[j][i]=1; int ans=0; memset(mark,sizeof(mark)); for(int i=1;i<=n;i++) { memset(vist,sizeof(vist)); ans+=DFS(i); } printf("%d\n",ans/2); } }
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