
解析:
想了好久,其实只需要输出 k-3 个1,然后问题就转换成为eazy版本了
eazy版本:1200*C1. k-LCM (easy version)(找规律)_陈进士学习的博客-CSDN博客
- #include
- using namespace std;
- int t,n,k;
- signed main(){
- scanf("%d",&t);
- while(t--){
- scanf("%d%d",&n,&k);
- for(int i=1;i<=k-3;i++) printf("1 ");
- n-=k-3;
- if(n%2==0){
- int x=n/2;
- if(x%2==0) printf("%d %d %d\n",x,x/2,x/2);
- else printf("%d %d %d\n",x-1,x-1,2);
- }
- else printf("%d %d %d\n",n/2,n/2,1);
- }
- return 0;
- }