欢乐纪中某A组赛【2019.1.19】

版权声明：原创，未经作者允许禁止转载 https://blog.csdn.net/Mr_wuyongcong/article/details/86553634

前言

因为 $B$有一堆(两道)题都做过，于是就来做A组了。

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成绩

$Rank$是有算别人的

$Rank$	$Person$	$Score$	$A$	$B$	$C$
$3$	$2017myself$	$210$	$100$	$100$	$10$
$22$	$2017lrz$	$100$	$0$	$100$	$0$
$32$	$2017xxy$	$50$	$0$	$50$	$0$
$36$	$2017zyc$	$20$	$0$	$20$	$0$
$39$	$2017hzb$	$0$	$0$	$0$	$0$
$39$	$2017lw$	$0$	$0$	$0$	$0$
$39$	$2017hjq$	$0$	$0$	$0$	$0$
$39$	$2017xjq$	$0$	$0$	$0$	$0$

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正题

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$T1:jzoj2941-$贿赂【数学期望 $,dfs$】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86553257

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$T2: jzoj2940-$生成输入数据【最小生成树 $,$并查集】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86553318

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$T3:jzoj3771-$小Z的烦恼【高精度,数学】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86553496

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$some\ of\ code$

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T2 对拍

#include<cstdio>
#include<cstring>
using namespace std;
int n,m,s,t,pow[26];
bool v[1000010];
int main()
{
	scanf("%d",&t);
	pow[1]=1;
	for(int i=2;i<=25;i++)
		pow[i]=pow[i-1]*2;
	while(t--)
	{
		memset(v,0,sizeof(v));
		s=0;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n/m;i++)
	  	if(!v[i]&&i*pow[m]<=n)
	  	{
	  	  	s++;
	  	  	int j=i,k=1;
		  	while(j<=n&&k<=m){
		  	  	v[j]=true;
		  	  	j*=2;k++;
		  	}
	  	}
		printf("%d\n",s);
	}
}

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尾声