前言

写完$T1$T1和$T2$T2颓了半天

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成绩

$Rank$Rank是有算别人的

$Rank$Rank	$Person$Person	$Score$Score	$A$A	$B$B	$C$C
$2$2	$2017myself$2017myself	$230$230	$100$100	$100$100	$30$30
$4$4	$2017xxy$2017xxy	$200$200	$100$100	$100$100	$0$0
$11$11	$2017zyc$2017zyc	$160$160	$100$100	$30$30	$30$30
$11$11	$2017lrz$2017lrz	$160$160	$100$100	$30$30	$30$30
$22$22	$2017xjq$2017xjq	$130$130	$100$100	$30$30	$0$0
$22$22	$2017hzb$2017hzb	$130$130	$100$100	$30$30	$0$0
$69$69	$2017lw$2017lw	$90$90	$60$60	$0$0	$30$30
$88$88	$2017hjq$2017hjq	$30$30	$0$0	$30$30	$0$0

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正题

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$T1:jzoj1273-$T1:jzoj1273−袁绍的刁难【进制转换】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86666724

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$T2:jzoj5230-$T2:jzoj5230−队伍统计【状压$dp$dp】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86666765

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$T3:jzoj5231-$T3:jzoj5231−序列问题【分治】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86666828

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$some\ of\ code$some of code

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T3 30分code

#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const ll XJQ=1e9+7;
ll n,a[100001],g,f,ans;
int main()
{
	freopen("seq.in","r",stdin);
	freopen("seq.out","w",stdout);
	scanf("%lld",&n);
	for(ll i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	for(ll i=1;i<=n;i++){
		f=0;g=1e18;
		for(ll j=i;j<=n;j++)
		{
			f=max(f,a[j]);
			g=min(g,a[j]);
			(ans+=f*g%XJQ)%=XJQ;
		}
	}
	printf("%lld",ans);
}

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总结

看题估分(100+60+30=190)

做题估分(100+100+30=230)

实际得分(100+100+30=230)居然一样?

T1这不是水题吗？切掉。T2开始不会，让去看T3也不会。然后十分绝望。分析一下T2数据范围，然后发现状压可以水60，结果发现其实是可以A掉。T3暴力30，一点思路也没有QVQ。

还好，和实际实力差不多

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尾声

某雷dalao的总结回复(上香的那个就是我)