题意：给你n和k，然后让你找出n个数使得gcd（a1,a2）+gcd(a3,a4)+......的和等于k；
思路：如果n为奇数，让前n-3个数的相邻两个数都为1，n-2和n-1两个数gcd为k-ans；ans为前n-3个数的和。为偶数的话，让前n-2个数的相邻两个数都为1，n和n-1两个数gcd为k-ans；
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <vector> 5 #include <algorithm> 6 #define maxn 10000 7 #define ll long long 8 using namespace std; 9 10 11 int n,k; 12 vector<int>q; 13 14 int main() 15 { 16 scanf("%d%d",&n,&k); 17 if((n%2==0&&(n/2)>k)||(n%2!=0&&(n-1)/2>k)) 18 { 19 printf("-1 "); 20 } 21 else 22 { 23 if(n==1&&k==0) 24 { 25 printf("1 "); 26 return 0; 27 } 28 else if(n==1&&k>0) 29 { 30 printf("-1 "); 31 return 0; 32 } 33 if(n%2==0) 34 { 35 int ans=0; 36 ll c=1000000000; 37 for(int i=1; i<=n-2; i+=2) 38 { 39 printf("%lld %lld ",c,c-1); 40 c-=2; 41 ans++; 42 } 43 printf("%lld %lld ",(ll)(k-ans),(ll)((k-ans)*2)); 44 } 45 else 46 { 47 int ans1=0; 48 ll c1=1000000000; 49 for(int i=1; i<n-2; i+=2) 50 { 51 printf("%lld %lld ",c1,c1-1); 52 c1-=2; 53 ans1++; 54 } 55 printf("%lld %lld",(ll)(k-ans1),(ll)(k-ans1)*2); 56 printf(" %lld ",c1); 57 } 58 } 59 return 0; 60 }