The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.
Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.
把质数当成节点，如果两数互相连接后仍为质数，则连一边，那么这个问题就转化成为在图中找出大小为k的完全子图。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 | #include<iostream> #include<cmath> #include<vector> #include<fstream> using namespace std; bool is_prime(int n); int concatNum(int a, int b); const int N=10000; int main() { ofstream out; out.open("p60.txt"); vector<int> primes; primes.push_back(2); for (int i = 3; i < N; i+=2) { if (is_prime(i)) primes.push_back(i); } for(int i=0; i<primes.size(); i++) { for (int j=0; j<i; j++) { if (is_prime(concatNum(primes[i], primes[j])) && is_prime(concatNum(primes[j], primes[i])) ) { out << primes[i] << "," << primes[j] << endl; } } } return 0; } bool is_prime(int n) { if ( n == 2) return true; if (n < 2) return false; if ( ! (n % 2) ) return false; for (int i=3; i<=sqrt(n); i+=2) { if (n % i == 0) return false; } return true; } int concatNum(int a, int b) { int res = a * pow((double) 10, (int) log10(b) +1) + b; return res; } |
先生成边列表文件，再用igraph包，找出完全子图。
1 2 3 4 5 6 | require(igraph) p <- read.csv("p60.txt", header=F) g <- graph.data.frame(p, directed=FALSE) subg <- cliques(g, min=5, max=5) res <- sapply(subg, function(i) as.numeric(V(g)$name[i])) min(colSums(res)) |