[开发工具]洛谷多校训练第 1 轮

T B
题面：

idea:这题一开始就知道是类似lqb里面的日期问题，但是当年准备lqb的时候并没有学习日期公式（就是蔡勒公式），只能写了一个没那么暴力的暴力程序。思路是先用十亿减去现在拥有的的钱数，然后一个月一个月地递推过去。

ACcode：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<map>
#define LL long long
#define INF 0x3f3f3f3f
#define N 20000000
using namespace std;

LL sum,money,t,n,year,month,day,l,r,tot;
LL mon[20];

LL getdays()
{
	LL l = 1,r = 160000;
	while( l<r )
	{
		LL mid = (l+r)/2;
		LL h = (mid*mid + mid)/2;
		if( h>=tot )
		{
			r = mid;
		}
		else l = mid+1;
	}
	return l;
}

bool judge( LL x )
{
	if( (x%4==0&&x%100!=0) || x%400==0 ) return true;
	else return false;
}

void getday(LL x,LL y,LL z,LL h)
{
	LL sheng,ydays;
	while(1)
	{
		if( y!=2 ) ydays = mon[y];
		else
		{
			if( judge(x)==true ) ydays = 29;
			else ydays = 28;
		}
		sheng = ydays - z;
		if( sheng >= h ) 
		{
			year = x;
			month = y;
			day = z;
			sum = h;
			return ;
		}
		h -= sheng;
		z = 0;
		y++;
		if( y>12 )
		{
			x++;
			y = 1;
		}
	}
}

void work()
{
	tot = 1000000000 - money;
	sum = getdays();
	getday(year,month,day,sum);
	printf("%lld %lld %lld\n",year,month,day+sum);
}

int main()
{
	mon[1] = 31;
	mon[2] = 28;
	mon[3] = 31;
	mon[4] = 30;
	mon[5] = 31;
	mon[6] = 30;
	mon[7] = mon[8] = 31;
	mon[9] = 30;
	mon[10] = 31;
	mon[11] = 30;
	mon[12] = 31;
	cin>>t;
	while( t-- )
	{
		scanf("%lld %lld %lld %lld",&money,&year,&month,&day);
		work();
	}
	return 0;
}

T H
题面：

idea:就是哥德巴赫猜想，给出一个小于1e9的数字，让你拆开成两个质数的和。（其实一亿内的质数挺少的）我们只需要晒出1到10000的质数，然后对每个n，减去筛出的质数后再判断是否是质数就好.O(√n log2 n)

ACcode:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<map>
#define LL long long
#define INF 0x3f3f3f3f
#define N 20000000
using namespace std;

LL pr[N],tot = 0;
int pd[N];
//map<LL,int> pd;
int sum = 0;
void get_prime()
{
	for(LL i=2;i<=100000;i++)
	{
		if( pd[i]!=1 )
		{
			tot++;
			pr[tot] = i;
		}
		for(int j=1;j<=tot;j++)
		{
			if( i*pr[j]>100000 ) break;
			pd[ i*pr[j] ] = 1;
			if( i%pr[j]==0 ) break;
		}
	}
}

LL n;
bool judge(LL x)
{
	LL y = sqrt(x);
	for(LL i=2;i<=y;i++)
	{
		if( x%i==0 ) return false;
	}
	return true;
}
int main()
{
//	freopen("out.txt","w",stdout);
	get_prime();
	cin>>n;
	for(int j=1;j<=tot;j++)
	{
		LL t = n-pr[j];
		if( judge( t )==true )
		{
			cout<<pr[j]<<" "<<t;
			return 0;
		}
	}	
	return 0;
}