ARC102 C~D

C:

枚举中间点，计算两边点差值，把个数乘起来即可

#include<iostream>

#include<cstdio>

#include<algorithm>

using namespace std;

const int N=200005;

long long n,k,ans;

int main()

{

	scanf("%lld%lld",&n,&k);

	for(long long i=1;i<=n;i++)

		if(2*(abs(k-i)%k)%k==0)

			ans+=1ll*((n-(abs(k-i)%k))/k+(abs(k-i)%k!=0))*((n-(abs(k-i)%k))/k+(abs(k-i)%k!=0));//,cerr<<i<<" "<<(n-(abs(k-i)%k))/k<<endl;

	printf("%lld\n",ans);

	return 0;

}

D：

#include<iostream>

#include<cstdio>

using namespace std;

const int N=20005;

long long n,k,w,k1,t,cnt,p,p1[N],kl[N],l[N],r[N],v[N];

int main()

{

	scanf("%d",&n);

	k=1;

	if(n==1)

	{

		printf("2 1\n1 2 0\n");

		return 0;

	}

	t=1;

	while(t*2<=n)

	{

		l[++cnt]=k;

		r[cnt]=k+1;

		v[cnt]=0;

		l[++cnt]=k;

		r[cnt]=k+1;

		v[cnt]=t;

		k++;

		t=t*2;

	}

	k1=k;

	n=n-t;

	w=t;

	while(n>0)

	{

		t=1;

		k=1;

		while(t*2<=n)

			t=t*2,k++;

		l[++cnt]=k;

		r[cnt]=k1;

		v[cnt]=w;

		w=w+t;

		n=n-t;

	}

	printf("%d %d\n",k1,cnt);

	for(int i=1;i<=cnt;i++)

		printf("%d %d %d\n",l[i],r[i],v[i]);

	return 0;

}