前言

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成绩

$Rank$是有算别人的

$Rank$	$Person$	$Score$	$A$	$B$	$C$
$2$	$2017myself$	$210$	$100$	$100$	$10$
$3$	$2017zyc$	$200$	$100$	$100$	$0$
$3$	$2017xjq$	$200$	$100$	$100$	$0$
$3$	$2017xxy$	$200$	$100$	$100$	$0$
$10$	$2017lw$	$130$	$100$	$30$	$0$
$14$	$2017hjq$	$60$	$60$	$0$	$0$
$17$	$2017hzb$	$30$	$30$	$0$	$0$
$18$	$2017lrz$	$20$	$0$	$20$	$0$

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正题

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$T1:jzoj3054-$祖孙询问【$LCA$】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/88359248

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$T2:jzoj3055-$比赛【数学$,$统计】

连接:
https://blog.csdn.net/Mr_wuyongcong/article/details/88359658

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$T3:jzoj3056-$数字【数位$dp,$统计$,$容斥】

连接:
https://blog.csdn.net/Mr_wuyongcong/article/details/88361756

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$someofcode$

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T2 暴力

#include<cstdio>
#define ll long long
using namespace std;
ll n,a[50010],b[50010],ans1,ans2;
int main()
{scanf("%lld",&n);for(ll i=1;i<=n;i++)scanf("%lld",&a[i]);for(ll i=1;i<=n;i++)scanf("%lld",&b[i]);for(ll i=1;i<=n;i++)for(ll j=1;j<=n;j++){ll w=(a[i]-b[j])*(a[i]-b[j]);if(a[i]>b[j]) ans1+=w;else ans2+=w;}printf("%.1lf",(double)(ans1-ans2)/n);
}

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T3 暴力

#include<cstdio>
#include<cstring>
using namespace std;
const int N=1010,XJQ=999983;
int n,Z,ans,m,a[12],c[12];
char s[12];
void dfs(int x,int ys,int ds)
{if(x==2*n){if(ys!=0&&ds!=0)return;//for(int i=0;i<2*n;i++)//  printf("%d",c[i]);//putchar('\n');ans=(ans+1)%XJQ;return;}for(int i=1;i<=m;i++){c[x]=a[i];dfs(x+1,ys+a[i]*((x%2)?1:-1),ds+a[i]*((x>=n)?1:-1));}
}
int main()
{scanf("%d",&n);scanf("%s",s+1);//double star=clock();m=strlen(s+1);for(int i=1;i<=m;i++)a[i]=s[i]-'0';dfs(0,0,0);printf("%d",ans);
}

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总结

看题预估分$100+30+20=150$
做题预估分$100+100+20=220$
实际分数$100+100+10=210$
$T1$明显$LCA$，直接切掉
$T2$开始不会，后来发现了全排列的性质，然后推一下公式就切掉了
之后死刚$T3$，然后公式全靠猜，结果10分

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尾声

没有尾声