地址：http://codeforces.com/contest/660/problem/A

題目：

A. Co-prime Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

You are given an array of?n?elements, you must make it a co-prime array in as few moves as possible.

In each move you can insert any positive integral number you want not greater than?109?in any place in the array.

An array is co-prime if any two adjacent numbers of it are co-prime.

In the number theory, two integers?a?and?b?are said to be co-prime if the only positive integer that divides both of them is?1.

Input

The first line contains integer?n?(1?≤?n?≤?1000) — the number of elements in the given array.

The second line contains?n?integers?ai?(1?≤?ai?≤?109) — the elements of the array?a.

Output

Print integer?k?on the first line — the least number of elements needed to add to the array?a?to make it co-prime.

The second line should contain?n?+?k?integers?aj?— the elements of the array?a?after adding?k?elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array?a?by adding?kelements to it.

If there are multiple answers you can print any one of them.

Example input 3
2 7 28 output 1
2 7 9 28

?思路：互質(zhì)是兩個(gè)數(shù)的最大公約數(shù)為1，即gcd（a，b）=1；

　　如果相鄰兩個(gè)數(shù)不是互質(zhì)的話，插個(gè)1就好了（比賽時(shí)沒想到1，插得是1—100內(nèi)某一質(zhì)數(shù)，（因?yàn)閤<10^9，所以100內(nèi)的質(zhì)數(shù)絕對可以，因?yàn)槌朔e大于最大取值范圍了）

　　代碼：

?

1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cmath> 5 #include <cstring> 6 #include <queue> 7 #include <stack> 8 #include <map> 9 #include <vector> 10 11 #define PI acos((double)-1) 12 #define E exp(double(1)) 13 using namespace std; 14 15 int a[1010]; 16 int b[5000]; 17 int prime[17] = { 0,7,13,19,23,31,37,41,43,47,53,59,61,67,71,73,79 }; 18 19 int is_coprime(int a, int b) 20 { 21 int t = 1; 22 while (t) 23 { 24 if (b == 1) 25 return 1; 26 t = a %b; 27 a = b; 28 b = t; 29 } 30 return 0; 31 } 32 33 int main(void) 34 { 35 int n, k; 36 while (scanf("%d", &n) == 1) 37 { 38 k = 1; 39 memset(b, 0, sizeof(b)); 40 for (int i = 1; i <= n; i++) 41 scanf("%d", &a[i]); 42 for (int i = 1; i<n; i++) 43 { 44 if (is_coprime(a[i], a[i + 1])) 45 { 46 b[k++] = a[i]; 47 } 48 else 49 { 50 b[k++] = a[i]; 51 for (int j = 1; j <= 16; j++) 52 if (is_coprime(a[i], prime[j]) && is_coprime(a[i + 1], prime[j])) 53 { 54 b[k++] = prime[j]; 55 break; 56 } 57 } 58 } 59 b[k] = a[n]; 60 cout << k - n << endl; 61 for (int i = 1; i<k; i++) 62 printf("%d ", b[i]); 63 printf("%d\n", b[k]); 64 } 65 return 0; 66 } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/weeping/p/5371872.html

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