前言

你以为我是$jzoj$，其实我是$GMoj$哒

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成绩

$J$表示初中，$H$表示高中后面加的是几年级
上至$2$分$XJQ$,下至$200ZZY$都有我们$SSL$的人(滑稽)
|$Rank$|$Person$|$Score$|$A$|$B$|$C$|

$Rank$	$Person$	$Score$	$A$	$B$	$C$
$2$	$(J-3)ZZY$	$200$	$100$	$100$	$0$
$15$	$(J-3)WHF$	$104$	$16$	$88$	$0$
$20$	$(J-2)ZYC$	$96$	$0$	$86$	$10$
$22$	$(J-2)WYC$	$90$	$4$	$86$	$0$
$27$	$(J-2)HJQ$	$84$	$0$	$84$	$0$
$34$	$(J-2)LRZ$	$52$	$8$	$44$	$0$
$36$	$(J-2)XXY$	$51$	$51$	$0$	$0$
$40$	$(J-3)QYH$	$38$	$2$	$34$	$2$
$57$	$(J-2)HZB$	$14$	$12$	$2$	$0$
$61$	$(J-2)LW$	$4$	$2$	$2$	$0$
$63$	$(J-2)XJQ$	$2$	$0$	$2$	$0$

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正题

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$T1:jzoj3318-[BOI2013]Brunhilda$的生日【数论】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/95064511

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$T2:jzoj3319-[BOI2013]$雪地踪迹【$bfs$】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/95069822

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$T3:Loj2687,jzoj3320-$文本编辑器【线头$dp$】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/95077052

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$someofcode$

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$T14$分$code$

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define check(x,y) ((x%y==0)?2147483647:x%y)
#define tr(x) (x+1)%m
#define cot(x,y) ((x%y==0)?(2147483647):(int(x/y)*y))
using namespace std;
const int N=10001;
int m,n,a[N],f[N],Q,mod[N],flag,pord;
int main()
{scanf("%d%d",&m,&Q);pord=1;for(int i=0;i<m;i++){scanf("%d",&a[i]);pord=pord*a[i];if(pord>=N) flag=1,pord=0;}memset(f,0x3f,sizeof(f));f[0]=0;for(int i=1;i<=N;i++){if(!flag&&i>=pord) break;for(int j=1;j<m;j++)if(i%a[j]>i%a[mod[i]])mod[i]=j;f[i]=f[cot(i,a[mod[i]])]+1;}while(Q--){scanf("%d",&n);if(!flag&&n>=pord) printf("oo\n");else printf("%d\n",f[n]);}
}

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$T286$分$code$

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=4100;
//int show[N][N];
int n,m,ans,v[N][N];
int qx[N*N],qy[N*N],l=1,r;
char c[N][N],z;
void dfs(int x,int y,char pos)
{if(x<1||y<1||x>n||y>m) return;if(c[x][y]!=pos||v[x][y]) return;v[x][y]=1;qx[++r]=x;qy[r]=y;dfs(x+1,y,pos);dfs(x,y+1,pos);dfs(x-1,y,pos);dfs(x,y-1,pos);
}
int main()
{scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)cin>>c[i][j];dfs(1,1,c[1][1]);z=c[1][1];while(l<=r){ans++;int R=r;if(z=='F') z='R';else z='F';for(int i=l;i<=R;i++){
//			show[qx[i]][qy[i]]=1;dfs(qx[i]-1,qy[i],z);dfs(qx[i]+1,qy[i],z);dfs(qx[i],qy[i]-1,z);dfs(qx[i],qy[i]+1,z);}
//		for(int i=1;i<=n;i++){
//		  for(int j=1;j<=m;j++)
//			printf("%c",show[i][j]*(z=='F'?'R':'F'));
//		  putchar('\n');
//		}
//		putchar('\n');l=R+1;}printf("%d",ans);
}

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总结

好烦今天我好菜，$T2$虽然一眼正解谁知$linux$也会爆栈，然后$T1$死刚半天结果是错误的方法。

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$$\text{;(}$$

最后的最后
$$\begin{gathered} \begin{array}{c}{我}_{T{M}^{{依}^{旧}}}^{{蔡}_{{的}_{{一}_{{匹}_{呀}}}}}\end{array} \\ \text{鸡你太美} \end{gathered}$$