Educational Codeforces Round 11A. Co-prime Array 数学

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

题目：

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：互质是两个数的最大公约数为1，即gcd（a，b）=1；

　　如果相邻两个数不是互质的话，插个1就好了（比赛时没想到1，插得是1—100内某一质数，（因为x<10^9，所以100内的质数绝对可以，因为乘积大于最大取值范围了）

　　代码：

 

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

 

转载于:https://www.cnblogs.com/weeping/p/5371872.html

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