Cena以及Lemon SpecialJudge的姿势
Jan 23, 2015 03:47:09 PM
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;
}
[开新坑]对于后缀自动机的一些理解
Jan 12, 2015 03:44:00 PM
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.
2015.01.04集训总结
Jan 04, 2015 08:04:53 PM
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~~~
JDFZOJ上的USACO月赛题目目前存在的BUG
Dec 23, 2014 09:59:22 PM
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