2018多校 第一场

A

给出一个数n，找到3个数a,b,c，a,b,c均能整除n，使得n = a+b+c且a*b*c最大，找不到a b c就输出-1

令 x = n/a y = n/b z = n/c 公式同时除以n--->1 = 1/x+1/y+1/z -- >3个解 （3，3，3）（2，4，4），（2，3，6）

当n%3 == 0--> x*y*z = n^3/27

n%4 == 0 -->x*y*z = n^3/32

n%6 == 0 -->x*y*z = n^3/36<n^3/27 = n^3/27

其他情况为-1

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

int main(){
	int t;
	long long n;
	scanf("%d",&t);
	while(t--){
		scanf("%lld",&n);
		if(n % 3 == 0)
		printf("%lld\n",n*n*n/27);
		else if(n % 4 == 0)
			printf("%lld\n",n*n*n/32);
		else
		    cout<<"-1\n";
		}
	return 0;
}

C

给出3*n个不共线的点，组合出n个三角形，求n个三角形的三个坐标的顺序

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

struct node{
	int x,y,num;
	bool operator< (const node& a) const{
	    return x<a.x || x==a.x&&y<a.y;
	}
}a[300006];


int main(){
	int t,n;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i = 1;i<=3*n;++i){
			scanf("%d%d",&a[i].x,&a[i].y);
			a[i].num = i;
		}
		sort(a+1,a+1+3*n);
		int j = 1;
		for(int i = 1;i<=n;++i){
			cout<<a[j].num<<" "<<a[j+1].num<<" "<<a[j+2].num<<endl;
			j += 3;
		}
	}
	return 0;
} 

Chiaki Sequence Revisited 找规律，a1 = a2 = 1,an = a n-an-1 + a n-1-an-2

#include<iostream>
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;

LL n;
bool judge(LL mid,LL n){
    LL ans = 0;
    while(mid > 0)
       ans += mid,mid >>= 1;
    return ans >= n; 
}
const LL inv2 = 500000004;

const int maxn = 1e6;
LL a[maxn];
LL an[maxn];
LL  F(LL n){

    if(n < 200){
         return a[n];
     }
     n--;

     LL l = 1,r = n;
     while(r >= l){
         LL mid = l+((r-l)>>1);
         if(!judge(mid,n))
            l = mid+1;
         else 
            r = mid-1;
     }
     LL ans = 0;
     LL num = 0;
     while( r > 0)
       num += r, r >>=1;
     num = n-num;
     ans = (num%mod*(l%mod))%mod;
     l--;
     LL tmp = 1;
     while(1){
          LL t = l/tmp;
          if(t & 1)
             t = t%mod*(((t+1)/2)%mod)%mod;
          else 
             t = (t/2)%mod*((t+1)%mod)%mod;
         ans = (ans+(t*(tmp%mod)%mod))%mod; 
         tmp <<= 1;
         if(tmp > l)
           break;
     }
    return (ans+1)%mod;
} 
int main(void)
{  
  a[1] =a[2]= 1;
  for(int i = 3;i < maxn; ++i)
     a[i] = (a[i-a[i-1]]+a[i-1-a[i-2]])%mod;

  for(int i = 1;i < maxn;++i)
     a[i] = (a[i] +a[i-1])%mod;


  int T;
  scanf("%d",&T);
  for(int i = 1;i <= T; ++i){
     scanf("%lld",&n);
     LL ans = F(n);
     printf("%lld\n",ans);
  } 


   return 0;
}