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[弃疗]直播玩提价答案题

2016-06-07 23:01:09 By absi2011

发现交互题不会做了..

还是提交答案题靠谱..

http://uoj.ac/problem/116

首先先手写个checker..

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<math.h>
#include<string>
#include<time.h>
#include<bitset>
#include<vector>
#include<memory>
#include<utility>
#include<stdio.h>
#include<sstream>
#include<fstream>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1<<23];
stringstream s;
ofstream fout;
int * get(int check_tag=0)
{
    static int operating[15];
    static int now=0;
    int x;
    char oper;
    s>>oper;
    s>>x;
    if (oper=='@')
    {
        if (check_tag)
        {
            printf("RuntimeError : Inputing/Equalling a constant\n");
            exit(0);
        }
        fout<<x<<" ";
        if (now==15) now=0;
        operating[now++]=x;
        return &operating[now-1];
    }
    else if (oper=='$')
    {
        if (x>=(1<<23))
        {
            printf("RuntimeError : Trying to visit a[%d]\n",x);
            exit(0);
        }
        else
        {
            fout<<"a["<<x<<"] ";
            return a+x;
        }
    }
    else
    {
        if (x>=(1<<23))
        {
            printf("RuntimeError : Trying to visit a[a[%d]]\n",x);
            exit(0);
        }
        else if (a[x]>=(1<<23))
        {
            printf("RuntimeError : Trying to visit a[a[%d]] (a[%d])\n",x,a[x]);
            exit(0);
        }
        else
        {
            fout<<"a[a["<<x<<"]] ";
            return a+a[x];
        }
    }
}
string line[100005];
char t[100005];
int main()
{
    ifstream fin;
    fin.open("lock1.in");
    fout.open("log.txt");
    int now=0;
    int n=1;
    for (;;)
    {
        fin.get(t,100000);
        if (t[0]=='\0') break;
        line[n++]=t;
        fin.get();
    }
    now=1;
    for (;;)
    {
        string oper;
        if (now==-1)
        {
            printf("\n\n\n");
            printf("Program Exit successfully");
            exit(1);
        }
        if ((now<-1)||(now>=n)||(now==0))
        {
            printf("RuntimeError : Trying to get %d-line",now);
            exit(0);
        }
        s.clear();
        s<<line[now];
        s>>oper;
        fout<<now<<endl;
        if (oper=="=")
        {
            int * t1=get(1);
            fout<<"=";
            int * t2=get();
            (*t1)=(*t2);
            fout<<" Value = "<<(*t2);
        }
        else if (oper=="geti")
        {
            int * t1=get(1);
            fout<<".input(int)";
            if (scanf("%d",t1)!=1)
            {
                printf("RuntimeErorr: Reading Failed (%%d)");
                exit(0);
            }
        }
        else if (oper=="getc")
        {
            int * t1=get(1);
            char tx;
            fout<<".input(char)";
            if (scanf("%c",&tx)!=1)
            {
                printf("RuntimeErorr: Reading Failed (%%c)");
                exit(0);
            }
            (*t1)=tx;
        }
        else if (oper=="puti")
        {
            int * t1=get();
            fout<<".print(int)";
            printf("%d",(*t1));
        }
        else if (oper=="putc")
        {
            int * t1=get();
            fout<<".print(char)";
            printf("%c",(*t1));
        }
        else if (oper=="if")
        {
            int * t1=get();
            int * t2=get();
            fout<<"Goto "<<(*t2)<<"?";
            if ((*t1)!=0) now=(*t2)-1;
            if ((*t1)!=0) fout<<" Yes";
        }
        else if (oper=="+")
        {
            int * t1=get();
            fout<<"+";
            int * t2=get();
            fout<<"-->";
            int * t3=get(1);
            (*t3)=(*t1)+(*t2);
            fout<<" Value = "<<(*t3);
        }
        else if (oper=="-")
        {
            int * t1=get();
            fout<<"-";
            int * t2=get();
            fout<<"-->";
            int * t3=get(1);
            (*t3)=(*t1)-(*t2);
            fout<<" Value = "<<(*t3);
        }
        else if (oper=="==")
        {
            int * t1=get();
            fout<<"==";
            int * t2=get();
            fout<<"-->";
            int * t3=get(1);
            (*t3)=((*t1)==(*t2));
            fout<<" Value = "<<(*t3);
        }
        else if (oper=="*")
        {
            int * t1=get();
            fout<<"*";
            int * t2=get();
            fout<<"-->";
            int * t3=get(1);
            (*t3)=(*t1)*(*t2);
            fout<<" Value = "<<(*t3);
        }
        else if (oper=="/")
        {
            int * t1=get();
            fout<<"/";
            int * t2=get();
            fout<<"-->";
            int * t3=get(1);
            if ((*t2)==0)
            {
                printf("RuntimeErorr : %d / 0 = ?",(*t1));
            }
            (*t3)=(*t1)/(*t2);
            fout<<" Value = "<<(*t3);
        }
        else if (oper=="<")
        {
            int * t1=get();
            fout<<"/";
            int * t2=get();
            fout<<"-->";
            int * t3=get(1);
            (*t3)=(*t1)<(*t2);
            fout<<" Value = "<<(*t3);
        }
        now++;
        fout<<endl;
    }
    return 0;
}

Test 1

写好checker就可以做题啦!

第一个输入点数字进去发现都会Too yaung..

始终在思考是yaung故意拼错还是什么...然后发现可能故意可能出题人手滑了..

反正..根据log.txt,大概发现是你输入的东西和23333333比较,如果一样就输出8个ok...

好的..过了

Test 2

第二个点好像是让你输入10个数字

输入第一个是1,发现输出了ko.....

输入了个0就给ok了...

我输入了10个0就全ok了....

果然是checker写错了我就知道不会那么简单..

改过以后看下log..


a[5] +0 -->a[0]  Value = 8388605
a[a[0]] =0  Value = 0
1 2314 Goto 2314? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -10
a[0] +a[5] -->a[0]  Value = 8388595
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = 7)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 1
a[a[0]] -1 -->a[0]  Value = 0
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -9
a[0] +a[5] -->a[0]  Value = 8388596
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = 47)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 2
a[a[0]] -1 -->a[0]  Value = 1
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -8
a[0] +a[5] -->a[0]  Value = 8388597
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = 1965)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 3
a[a[0]] -1 -->a[0]  Value = 2
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -7
a[0] +a[5] -->a[0]  Value = 8388598
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = 1915)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 4
a[a[0]] -1 -->a[0]  Value = 3
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -6
a[0] +a[5] -->a[0]  Value = 8388599
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = -2551)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 5
a[a[0]] -1 -->a[0]  Value = 4
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -5
a[0] +a[5] -->a[0]  Value = 8388600
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = -1646938625)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 6
a[a[0]] -1 -->a[0]  Value = 5
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -4
a[0] +a[5] -->a[0]  Value = 8388601
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = -322)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 7
a[a[0]] -1 -->a[0]  Value = 6
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -3
a[0] +a[5] -->a[0]  Value = 8388602
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = -167542220)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 8
a[a[0]] -1 -->a[0]  Value = 7
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -2
a[0] +a[5] -->a[0]  Value = 8388603
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = 4346926)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 9
a[a[0]] -1 -->a[0]  Value = 8
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 1
a[0] 2294 Goto 2294? Yes
a[0] .input(int)
a[a[4]] =a[0]  Value = 0
a[4] -1 -->a[4]  Value = 8388593
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +-10 -->a[0]  Value = -1
a[0] +a[5] -->a[0]  Value = 8388604
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 0
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = 1531256182)
a[0] 2308 Goto 2308?
107 
111 
10 
1 2311 Goto 2311? Yes
a[5] +0 -->a[0]  Value = 8388605
a[a[0]] +1 -->a[a[0]]  Value = 10
a[a[0]] -1 -->a[0]  Value = 9
a[5] +0 -->a[0]  Value = 8388605
a[a[4]] =a[0]  Value = 8388605
a[4] -1 -->a[4]  Value = 8388593
a[4] +1 -->a[4]  Value = 8388594
a[1] =a[a[4]]  Value = 8388605
a[a[1]] /10 -->a[0]  Value = 0
a[0] 2294 Goto 2294?
a[0] =0  Value = 0
a[4] +11 -->a[4]  Value = 8388605
a[4] =a[5]  Value = 8388605
a[4] +1 -->a[4]  Value = 8388606
a[5] =a[a[4]]  Value = 0
a[4] +1 -->a[4]  Value = 8388607
1 a[a[4]] Goto -1? Yes

这不就出来了么...

Test 3

lock3醚一样的让你输入一大堆东西..不是很懂怎么弄..

感觉输入全部被无视掉了..因为它让我反复输入..

好吧是输入20个数字...

应该是几个子问题,每个子问题有个ok/fail

task1: 输入完20个数 奖励: 1个ok

task2: 积为0 奖励: 3个ok

task3: 和为0 奖励: 3个ok

task4: 所有数里面没有0 奖励: 3个ok

所有数字均为-2147483648即可

Test 4

随便输入点东西

给了3个ok,并且说输入的太短了

二分下长度,长了会给4个ok

长度对了给5个,并且输出了No!

我觉得我可以去试试..看log

log上的判定很容易的让我判定出来了东西..

我发现第一个是97 != 112 所以我就把输入的字符给改了下

第一个字符对了,第二个错了..

之后就一直改到对了为止..

前面pr我以为是print结果竟然是primary..

primaryshcool!

然后后面的词..p开头的..pupil..?

过了

Test 5

输入10个字符串

我输入一个以后ctrl-C看看log里面的东西...

log大法好

a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 374 && *t2 = 25)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 374 && *t2 = 23)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 374 && *t2 = 11)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 274 && *t2 = 17)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 274 && *t2 = 15)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 318667 && *t2 = 14336)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 7216519 && *t2 = 205973)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 846609806 && *t2 = 43563464)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 11227352 && *t2 = 506369272)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 217741918 && *t2 = 214107364)

前几个就是..字符

后面的不知道怎么弄,不过..试试吧

374我输入的是ok

为啥ok = 374而ko = 274呢..

'o' = 111 'k' = 107 o : 14 k : 10 $10*26+14 = 274$

$14*26+10 = 374$

我觉得明确了..

我估计最后几个点我的答案绝对不是作者想要的2333

作者在取模..我又不是没事儿取模玩的人...

Test 6

lock6还是10个串?

突然感觉有checker就是好..log真心神器

这次是31进制的..execiting...

a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 0 && *t2 = 15)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 1 && *t2 = 13)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 2 && *t2 = 394136050)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 3 && *t2 = 951319991)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 4 && *t2 = 973533818)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 5 && *t2 = 796026279)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 6 && *t2 = 254508514)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 7 && *t2 = 740118828)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 9 && *t2 = 929913515)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 8 && *t2 = 208126208)

PS:t1是我输入的东西,所以会存在t1=9和t1=8反了的情况..我字母表背错啦

还有上面一句话莫名成链接了

然而...

Are you ready? Please enter 10 strings.
p
ok
n
ok
nxybfe
ok
bchdeeu
ok
bdaeyow
ok
|y~lgc
Expected a string consisting of a-z

就这么炸了..

...没办法咯..

这里是UOJ!取模的数字一定是998244353

去乖乖的加质数再取模吧

稍微改下5的程序即可..

lock6.cpp
#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<math.h>
#include<string>
#include<time.h>
#include<bitset>
#include<vector>
#include<memory>
#include<utility>
#include<stdio.h>
#include<sstream>
#include<fstream>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
/*
t2 = 25) (z)
t2 = 23) (x)
t2 = 11) (l)
t2 = 17) (r)
t2 = 15) (p)
t2 = 14336)
t2 = 205973)
t2 = 43563464)
t2 = 506369272)
t2 = 214107364)
*/
const int modo=998244353;
int main()
{
    freopen("lock6.out","w",stdout);
    int n=10;
    int i;
    for (i=0;i<n;i++)
    {
        long long x;
        cin>>x;
        long long p=1;
        long long tx=x;
        for (;;)
        {
            p=1;
            for (;;)
            {
                p*=31;
                if (p>x) break;
            }
            tx=x;
            for (;;)
            {
                p/=31;
                if (p==0) break;
                int t=tx/p;
                tx-=t*p;
                if (t>=26) break;
            }
            if (p==0) break;
            x+=998244353;
        }
        p=1;
        for (;;)
        {
            p*=31;
            if (p>x) break;
        }
        for (;;)
        {
            p/=31;
            if (p==0) break;
            int t=x/p;
            x-=t*p;
            if (t>=26) break;
            printf("%c",t+'a');
        }
        printf("\n");
    }
}

Test 7

然后....

第7个点开始有TLE的迹象了?

a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 1 && *t2 = 0)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 499122177 && *t2 = 1)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 332748118 && *t2 = 55555)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 748683265 && *t2 = 66666666)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 598946612 && *t2 = 234324325)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 166374059 && *t2 = 643522257)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 855638017 && *t2 = 865106053)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 873463809 && *t2 = 179441119)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 443664157 && *t2 = 864389574)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 299473306 && *t2 = 407615848)

感觉自己QAQ......

这tm是啥

0 : 0

1 : 1

2 : 499122177 这是啥数字啊

好吧是逆元什么的..

所以..这怎么办呢....

暴力大法好了

很快我意识到这复杂度是吃不消的gg...

然而想到$A^{-1^{-1}} = A$....没了

lock7.cpp
#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<math.h>
#include<string>
#include<time.h>
#include<bitset>
#include<vector>
#include<memory>
#include<utility>
#include<stdio.h>
#include<sstream>
#include<fstream>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int modo=998244353;
int power(int x,int y)
{
    if (y==0) return 1;
    int t=power(x,y/2);
    t=(long long)t*t%modo;
    if (y%2==1) t=(long long)t*x%modo;
    return t;
}
int a[15]={0,1,55555,66666666,234324325,643522257,865106053,179441119,864389574,407615848};
int main()
{
    freopen("lock7.out","w",stdout);
    int i;
    for (i=0;i<10;i++)
    {
        int t=power(a[i],modo-2);
        printf("%d\n",t);
    }
    return 0;
}

Test 8

又看到有77977这种进制的存在了...

不是很懂..输入的是数字

a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 1 && *t2 = 0)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 922932310 && *t2 = 1)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 792803569 && *t2 = 55555)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 194654562 && *t2 = 66666666)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 911030197 && *t2 = 234324325)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 591537984 && *t2 = 643522257)
a[1] ==a[a[0]] -->a[0]  Value = 0 ( *t1 = 170973641 && *t2 = 865106053)

不用看了肯定和上面的是同一组..数字..

现在的问题是.......前面那个*t1是什么鬼

输入x,则输出$x^{77977}$.....

取模的数字依旧...

然而这个我真的不会做啊.....

大概5s一个ok,一共100个ok大概要500s感觉不是很慌

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<math.h>
#include<string>
#include<time.h>
#include<bitset>
#include<vector>
#include<memory>
#include<utility>
#include<stdio.h>
#include<sstream>
#include<fstream>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int modo=998244353;
int power(int x,int y)
{
    if (y==0) return 1;
    int t=power(x,y/2);
    t=(long long)t*t%modo;
    if (y%2==1) t=(long long)t*x%modo;
    return t;
}
int a[15]={0,1,55555,66666666,234324325,643522257,865106053,179441119,864389574,407615848};
int ans[15];
int main()
{
    freopen("lock8.out","w",stdout);
    int i;
    for (i=0;i<modo;i++)
    {
        int t=power(i,77977);
        int j;
        for (j=0;j<10;j++)
        {
            if (t==a[j]) ans[j]=i;
        }
        if (i%10000000==0) fprintf(stderr,"ok\n");
    }
    for (i=0;i<10;i++)
    {
        printf("%d\n",ans[i]);
    }
    return 0;
}

Test 9(UnAC)

目前看懂了它在对10取模..

有趣了

前面俩OK基本是送的..只要你输入了就有

如果输入的数字全是0会TLE...

后面暂时不太懂

感觉是10进制的数字!

你要输入一个高精度数字...然后要让它的某个属性=0

这个属性是啥我也不知道....

已知:

1,输入0会TLE

2,输入n位数,该属性最大为n

3,输入1似乎会被直接退掉..

看代码感觉..是个高精度..除法?

虽然底下的东西不是很懂..

但是很多地方有醚一样的*10...

而且乘出来的结果只有0和10两种..

难道是高精度减法模拟除法?返回余数有几位不是0,如果全是0就输出8个Yes

难道是A mod B problem?

目标:找出一个不是1和本身的因数!

81135286724930827243695738730846577445642314805943搜了下$2~2^{32}-1$竟然都不行....我有种要报警的想法了

接着搜看int范围内有没有吧.....

我相信..根据test 10的提示..

这个数一定是x^3的样子...

果断写了个立方根没开出来..

Test 10(UnAC)

随便输入俩不是1并且..大于等于0的数字,输出一个ok

后面的骗不出来了...

第10个点应该是输入两个数字

然后输出1个ok

然后进行一次运算如果符合一个条件则给3个ok

然后再进行一次运算..

虽然我不是很会分解质因数!

但是当我输入81135286724930827243695738730846577445642314805943 81135286724930827243695738730846577445642314805943的时候

他给了我4个ok!

里面似乎是..高精度乘法?不是很懂什么鬼情况...

两个数乘起来并不是里面的那个大数字..不是很懂..

换了一些数是过了的..

真心不明白到底是啥情况..

还有50位数到底怎么找质因数..

最终得分 : 86

(通过搜索$wolframalpha$得到100分)

我弃疗了..

主要是因为Test 9那个整数我实在是不会分解了..

去看题解了..

题解链接在这里..http://mxh1999.blog.163.com/blog/static/247319071201542853335300/

你们好可怕竟然不认真做题用$醚$一样的办法把它给爆了..

我是不想说什么..

我只能认真的做出1~8个点了..

后面的点实在是sxbk..不想说什么

竟然后面两个点$醛$是这么利用爆栈乱搞的我服...

认输认输..

PS:据说可以分解..

第9个点:

http://www.wolframalpha.com/input/?source=nav&i=the+Divisors+of+81135286724930827243695738730846577445642314805943

$8873721024639708188763151 * 9143321781205657319947993 = 81135286724930827243695738730846577445642314805943$

第10个点:

http://www.wolframalpha.com/input/?source=nav&i=the+Divisors+of+35442943893941093802986826887594209913195827438699

$5364013213705248146033063 * 6607542241578206359887773 = 35442943893941093802986826887594209913195827438699$

得出结论:vfk真良心出题人...

把答案长度压缩了下成功到达rank 1..虽然好像没啥意义..http://uoj.ac/submission/76216

评论

QwX
前排
zhourunlong
前排！
shanquan2
前排！
WrongAnswer
第9个点正确姿势： 【首先我想知道它最多能输多少个数】 in： 5 1 1 1 1 1 out： ok ok 程序结束 【可以输入更多】 于是输入100，然后输入100个1 out： ok ok 程序结束 【可以输入更多】 于是输入200，然后…… 结果…… Runtime Error 【改小一点】 150……还是RE 120……还是RE 105……还是RE 【手动二分？】 于是输入了104，然后是104个1 out： ok ok 程序并没有结束！ 【所以可以继续玩咯？】 再输104，然后104个1 out： ok ok 还可以继续玩！ 再输104，然后104个1 out： ok ok 继续104…… 继续ok…… …… 好的，10个ok都搞到手了！
absi2011
@WrongAnswer 回复不能超过140个字节所以我在这里回复了.. 我把第九个点的程序的内容终于搞懂了是求a%b == 0然后a是给定的,我以为我能暴力2~$2^{32}-1$就够结果....并没有解.. 现在我在怀疑@vfleaking是不是做了什么手脚比如说丢了个质数上去 50位的数字怎么分解啊求教

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