Multiples of Length CodeForces - 1396A

原题链接
考察:构造
思路:
  对于n,如果每次x都能+-(n-1)的倍数,那么一定可以使x变成n的倍数.

\[x = n*x-(n-1)*x \]

\[x+(n-1)x = n*x \]

  由此这三步为:

1. 使a[1~n-1]变为n的倍数
2. 使a[n]变成0
3. 使a[1~n]变成0

Code

#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 100010;
int n;
LL a[N];
void solve()
{
	if(n==1)
	{
		printf("%d %d\n%lld\n",1,1,-a[1]);
		printf("%d %d\n0\n",1,1);
		printf("%d %d\n0\n",1,1);
		return;
	} 
	printf("%d %d\n",1,n-1);
	for(int i=1;i<n;i++)
	{
		printf("%lld ",(n-1)*a[i]);
		a[i]+=(n-1)*a[i];
	}
	printf("\n%d %d\n%lld\n",n,n,-a[n]);
	a[n] = 0;
	printf("%d %d\n",1,n); 
	for(int i=1;i<=n;i++)
	  printf("%lld ",-a[i]);
	printf("\n");
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	solve(); 
	return 0;
}