前言

心态爆炸

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成绩

$Rank$是有算别人的

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

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正题

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$T1:jzoj4235-$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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$T2:jzoj4226-A$【图论】

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

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$T3:jzoj4227-B$【$dp,$字符串】

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

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$T4:jzoj4228-C$【$dp$】

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

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$someofcode$

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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