这能算是普及?只是自己留一个存档.

有需要的就来自己脑补吧.

UPD:留一个lemon的对应参数说明：

Special Judge参数传送说明：
argv[1]: 标准输入文件
argv[2]: 选手输出文件
argv[3]: 标准输出文件
argv[4]: 本测试点满分
argv[5]: 分数输出文件（必须创建），仅一行，包含一个非负整数，表示得分。
argv[6]: 额外信息文件（可以不创建）

#include <cstdio>
#include <cstring>
#include <cstdlib>
FILE *fscore, *freport, *fstd, *fin, *fout;

inline void getScore(int x) {
	fprintf(fscore, "%d", x);
}

char s[2010], output[2010], tmp[1000000], *o = tmp;
inline int Judge() {
	fscanf(fin, "%s", s);
	int len = strlen(s);
	
	fscanf(fout, "%s", output);
	int len_out = strlen(output);
	
	int std_len;
	fscanf(fstd, "%d", &std_len);
	if (len_out != std_len)
		return 1;
	
	register int i, j, k, p, q;
	for(i = 0; i < len_out; ++i)
		if (output[i] != '$' && output[i] != '*' && !(output[i] >= 'a' && output[i] <= 'z'))
			return 2;
	if (output[len_out - 1] != '$')
		return 2;

	for(i = 0; i < len_out; ) {
		if (!(output[i] >= 'a' && output[i] <= 'z'))
			return 2;
		for(j = i; output[j] != '$'; ++j);
		for(k = i; output[k] != '*' && output[k] != '$'; ++k);
		for(p = k; p <= j; ++p)
			if (output[p] != '$' && output[p] != '*')
				return 2;
		for(p = 1; p <= j - k + 1; ++p)
			for(q = i; q < k; ++q)
				*o++ = output[q];
		i = j + 1;
	}
	
	int outputlen = o - tmp;
	if (outputlen != len)
		return 3;
	for(i = 0; i < len; ++i)
		if (tmp[i] != s[i])
			return 3;
	
	return 0;
}

int main(int argc, char *argv[]) {
	fscore = fopen("score.log", "w");
	freport = fopen("report.log", "w");
	
	fstd = fopen(argv[2], "r"); //测试点标准输出文件
	
	fin = fopen("compress.in", "r");
	fout = fopen("compress.out", "r"); //用户输入、输出文件
	
	if (!fout) {
		getScore(0);
		fprintf(freport, "Empty File!");
	}
	else {
		int res = Judge();
		if (res == 0)
			getScore(10), fprintf(freport, "Right Output!");
		else if (res == 1)
			getScore(0), fprintf(freport, "Wrong Length!");
		else if (res == 2)
			getScore(0), fprintf(freport, "Wrong Format!");
		else if (res == 3)
			getScore(0), fprintf(freport, "Wrong Change!");
	}
	
	fclose(fscore);
	fclose(freport);
	
	return 0;
}

#include<bits/stdc++.h>

FILE *fscore,*freport,*fstd,*fin,*fout;

inline void GetScore(int d){
	fprintf(fscore,"%d",d);
}
inline void End(){
	fclose(fscore),fclose(freport);
}

int d[1010],x[1000010],y[1000010];
inline void work(int a,int b){
	d[b]-=d[a],d[a]*=2;
}
int main(int argc,char*argv[]){
	fscore=fopen(argv[5],"w");//得分文件
	freport=fopen(argv[6],"w");//报告文件
	fstd=fopen(argv[3],"r");//标准输出
	fin=fopen(argv[1],"r");//标准输入
	fout=fopen(argv[2],"r");//用户输出
	
	int stdans;fscanf(fstd,"%d",&stdans);
	int yourans;
	if(fscanf(fout,"%d",&yourans)==0){
		fprintf(freport,"No output.");
		GetScore(0);
		End();return 0;
	}
	
	if(stdans==-1){
		if(yourans!=-1){
			fprintf(freport,"No solution but you output an ans");
			GetScore(0);
			End();return 0;
		}
		else{
			fprintf(freport,"OK"),GetScore(10);
			End();return 0;
		}
	}
	else{
		if(yourans==-1){
			fprintf(freport,"There exists some solutions but you output no ans");
			GetScore(0);
			End();return 0;
		}
		else if(yourans<0){
			fprintf(freport,"Wrong:T<0");
			GetScore(0);
			End();return 0;
		}
		else if(yourans>1000000){
			fprintf(freport,"Too many operates");
			GetScore(0);
			End();return 0;
		}
		else{
			int n;register int i;
			fscanf(fin,"%d",&n);
			for(i=1;i<=n;++i)fscanf(fin,"%d",&d[i]);

			for(i=1;i<=yourans;++i){
				if(fscanf(fout,"%d%d",&x[i],&y[i])!=2){
					fprintf(freport,"output Not enough");
					GetScore(0);
					End();return 0;
				}
			}
			
			int cash;
			if(fscanf(fout,"%d",&cash)==1){
				fprintf(freport,"Too many output");
				GetScore(0);
				End();return 0;
			}
			
			for(i=1;i<=yourans;++i){
				if(x[i]==y[i]){
					fprintf(freport,"Wrong at Line%d:x=y",i+1);
					GetScore(0);
					End();return 0;
				}
				else if(d[x[i]]>d[y[i]]){
					fprintf(freport,"Wrong at Line%d:a[x]>a[y]",i+1);
					GetScore(0);
					End();return 0;
				}
				else work(x[i],y[i]);
			}
			
			int last=0;
			for(i=1;i<=n;++i)if(d[i]>0)++last;
			if(last==2){
				fprintf(freport,"OK"),GetScore(10);
				End();return 0;
			}
			else{
				fprintf(freport,"yourlast=%d,and it's larger then 2",last),GetScore(0);
				End();return 0;
			}
		}
	}
	return 0;
}

感觉单纯地从"后缀自动机"的角度来入手并不是非常合理.因为我们懂得很多"后缀自动机"的性质,但却并不清楚"后缀自动机"在本质上是什么.

让我们从后缀树说起.

Task#1 Contest

面对三道傻逼题本沙茶妥妥爆零.大家一起喊出来我弱不弱,弱不弱?

T1题目大意:傻逼题,大家通过代码来猜测什么题目吧.(多带了一个\(log\)结果没有超时,反而由于空间问题被卡了2MB内存,艹!)

#include<cstdio>
#include<cstring>
#include<cctype>
#include<iostream>
#include<algorithm>
using namespace std;

#define N 100010
struct Node{
	int l,r,siz;
}S[N*20];int cnt;
int Newv(int last,int tl,int tr,int ins){
	int q=++cnt;S[q]=S[last],++S[q].siz;if(tl==tr)return q;
	int mid=(tl+tr)>>1;
	if(ins<=mid)S[q].l=Newv(S[last].l,tl,mid,ins);else S[q].r=Newv(S[last].r,mid+1,tr,ins);
	return q;
}
int Query(int q,int tl,int tr,int dl,int dr){
	if(dl<=tl&&tr<=dr)return S[q].siz;
	int mid=(tl+tr)>>1;
	if(dl>mid)return Query(S[q].r,mid+1,tr,dl,dr);
	else if(dr<=mid)return Query(S[q].l,tl,mid,dl,dr);
	else return Query(S[q].l,tl,mid,dl,mid)+Query(S[q].r,mid+1,tr,mid+1,dr);
}
struct Case{
	int x,y;
	bool operator<(const Case&B)const{return x<B.x;}
}C[N];
int p[N],q[N],root[N];
int main(){
	freopen("permutation.in","r",stdin),freopen("permutation.out","w",stdout);
	int n;cin>>n;register int i,j;
	for(i=1;i<=n;++i)scanf("%d",&p[i]),C[p[i]].x=i;for(i=1;i<=n;++i)scanf("%d",&q[i]),C[q[i]].y=i;
		sort(C+1,C+n+1);
	for(i=1;i<=n;++i)root[i]=Newv(root[i-1],1,n,C[i].y);
	int lans=0,a,b,c,d,aa,bb,cc,dd,q;scanf("%d",&q);
	while(q--){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		aa=(a-1+lans)%n+1,bb=(b-1+lans)%n+1,cc=(c-1+lans)%n+1,dd=(d-1+lans)%n+1;
		a=min(aa,bb),b=max(aa,bb),c=min(cc,dd),d=max(cc,dd);
		printf("%d\n",(lans=Query(root[b],1,n,c,d)-Query(root[a-1],1,n,c,d)+1)-1);
	}return 0;
	fclose(stdin),fclose(stdout);return 0;

}

T2:见下面的链接

http://wyfcyx.is-programmer.com/posts/75155.html

T3:见下面的链接

http://wyfcyx.is-programmer.com/posts/75156.html

Task#2 后缀数组及其应用

现在没时间,等我找到特别有意思的题目来做吧~~~

刷到哪里更到哪里。

Bronze:

Silver:

Gold:

2567:提交答案

2586:需要SPJ

2607:需要SPJ

2655:需要SPJ

2664&1013:需要SPJ