CodeForces - 660A Co-prime Array （模拟）

CodeForces - 660A Co-prime Array Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description 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 k elements to it. If there are multiple answers you can print any one of them. Sample Input Input 3 2 7 28 Output 1 2 7 9 28 Source Educational Codeforces Round 11 //题意： 给你一个有n个元素的序列，现在让你在序列中加入k个数（最少），使得这个序列任意的相邻的两个元素互质。输出这个最小的数k并将加入后的序列输出 //思路： 直接判断相邻两个元素是否互质（gcd(a,b)==1），不互质的话在中间加个1即可。 #include<stdio.h> #include<string.h> #include<math.h> #include<set> #include<map> #include<stack> #include<queue> #include<algorithm> #include<iostream> #define INF 0x3f3f3f3f #define ull unsigned long long #define ll long long #define IN __int64 #define N 100010 #define M 1000000007 using namespace std; int a[N],b[N]; int gcd(int a,int b) { return b?gcd(b,a%b):a; } int main() { int n,m,i,j,k; while(scanf("%d",&n)!=EOF) { for(i=0;i<n;i++) scanf("%d",&a[i]); k=0; for(i=0;i<n;i++) { if(i!=n-1) { if(gcd(a[i],a[i+1])==1) b[k++]=a[i]; else b[k++]=a[i],b[k++]=1; } } b[k++]=a[n-1]; printf("%d\n",k-n); for(i=0;i<k;i++) printf("%d ",b[i]); printf("\n"); } return 0; }