前言

心态爆炸

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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	$D$D
$3$3	$2017myself$2017myself	$190$190	$100$100	$50$50	$0$0	$40$40
$13$13	$2017zyc$2017zyc	$170$170	$70$70	$60$60	$0$0	$40$40
$13$13	$2017hzb$2017hzb	$170$170	$100$100	$70$70	$0$0	$0$0
$22$22	$2017xjq$2017xjq	$150$150	$100$100	$50$50	$0$0	$0$0
$33$33	$2017lrz$2017lrz	$130$130	$80$80	$50$50	$0$0	$0$0
$44$44	$2017lw$2017lw	$100$100	$40$40	$60$60	$0$0	$0$0
$57$57	$2017xxy$2017xxy	$90$90	$40$40	$50$50	$0$0	$0$0
$57$57	$2017hjq$2017hjq	$90$90	$40$40	$50$50	$0$0	$0$0

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正题

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$T1:jzoj4235-$T1:jzoj4235−序列【斐波那契数列】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86679575

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$T2:jzoj4226-A$T2:jzoj4226−A【图论】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86679695

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$T3:jzoj4227-B$T3:jzoj4227−B【$dp,$dp,字符串】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86679832

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$T4:jzoj4228-C$T4:jzoj4228−C【$dp$dp】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86679939

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$some\ of\ code$some of code

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T2 30分code

#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=100010;
ll n,m,k,wall[N],ans;
int main()
{
	scanf("%lld%lld%lld",&n,&m,&k);
	for(ll i=1;i<=m;i++){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		wall[x]++;wall[y]++;
	}
	sort(wall+1,wall+1+n);
	for(ll i=n;i>=2;i--)
		if(wall[i]<k){
			ll z=k-wall[i];
			wall[i]+=z;
			wall[i-1]+=z;
			ans+=z;
		}
	if(wall[1]!=k) ans+=k-wall[1];
	printf("%lld",ans);
}

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T4 40分code

#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
struct node{
	ll x,y;
}a[100];
ll n,v[100][4],ans;
void dfs(ll dep){
	if(dep>n){
		ans=(ans+1)%998244353;
		return;
	}
	if(!v[dep][0]){
		for(ll i=dep+1;i<=n;i++)
			if(a[i].x>a[dep].x&&a[i].y>a[dep].y) v[i][3]++;
		dfs(dep+1);
		for(ll i=dep+1;i<=n;i++)
			if(a[i].x>a[dep].x&&a[i].y>a[dep].y) v[i][3]--;
	}
	if(!v[dep][1]){
		for(ll i=dep+1;i<=n;i++)
			if(a[i].x>a[dep].x&&a[i].y<a[dep].y) v[i][3]++;
		dfs(dep+1);
		for(ll i=dep+1;i<=n;i++)
			if(a[i].x>a[dep].x&&a[i].y<a[dep].y) v[i][3]--;
	}
	if(!v[dep][2]){
		for(ll i=dep+1;i<=n;i++)
			if(a[i].x>a[dep].x&&a[i].y<a[dep].y) v[i][0]++;
			else if(a[i].x>a[dep].x) v[i][1]++;
		dfs(dep+1);
		for(ll i=dep+1;i<=n;i++)
			if(a[i].x>a[dep].x&&a[i].y<a[dep].y) v[i][0]--;
			else if(a[i].x>a[dep].x) v[i][1]--;
	}
	if(!v[dep][3])
		dfs(dep+1);
}
bool cmp(node x,node y){
	return x.x<y.x;
}
int main()
{
	scanf("%lld",&n);
	for(ll i=1;i<=n;i++)
		scanf("%lld%lld",&a[i].x,&a[i].y);
	sort(a+1,a+1+n,cmp);
	for(ll i=1;i<=n;i++){
		for(ll j=i+1;j<=n;j++){
			if(a[i].x==a[j].x&&a[i].y<a[j].y) v[i][0]=1,v[j][1]=1;
			if(a[i].x==a[j].x&&a[i].y>a[j].y) v[i][1]=1,v[j][0]=1;
			if(a[i].y==a[j].y&&a[i].x<a[j].x) v[i][2]=1,v[j][3]=1;
			if(a[i].y==a[j].y&&a[i].x>a[j].x) v[i][3]=1,v[j][2]=1;
		}
	}
	dfs(1);
	printf("%lld",ans);
}

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总结

期望:(30+100+0+30=160)
实际:(100+50+0+40=190)还多了

T1不会就去刚T2，感觉还行结果没有判断另一种情况。之后去看T3，看题看了半天，不会。之后看T4，写了个30分暴力，结果拿了40。之后T1写了个三十分暴力，然后竟然过了(woc???)。

赛后发现T1暴力就是可以过，因为斐波那契数列到50就够了

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尾声

今日说法