Digital informatization (enumerate assumptions first, and then see whether the conditions are met) (1089 werewolf kill - simple version)

question 1 The police caught him a,b,c,d Four suspected thieves , Only one of them was a thief .
In Interrogation ,a say ： I'm not a thief ;b say ：c It's a thief ;c say ： The thief must be d;d say ：c Wronged people .
It is now known that 3 People tell the truth , One person lies , Ask who the thief is ;
 * \ anaysis： Number the thief , Enumerate all people trying , Meet the conditions is a thief .

/** \ question 1  The police caught him  a,b,c,d  Four suspected thieves , Only one of them was a thief .
        In Interrogation ,a say ： I'm not a thief ;b say ：c It's a thief ;c say ： The thief must be d;d say ：c Wronged people .
 *      It is now known that 3 People tell the truth , One person lies , Ask who the thief is ;
 * \ anaysis： Number the thief , Enumerate all people trying , Meet the conditions is a thief .
 * \param
 * \return
 *
 */
/**<  */

/**
#include <iostream>

using namespace std;

int main()
{
    int thief='a';
    for(thief='a';thief<='d';++thief)
        if(((thief!='a')+(thief=='c')+(thief=='d')+(thief!='d'))==3)
            printf("%c is thief\n",thief);
    return 0;
}
*/

/** \ question 2:3 A teacher predicted a math contest , Their predictions are as follows ：
        A said ：a Won the first place ,b Got the third place ;
        B said ：c Won the first place ,d Got the fourth place ;
        C said ：d Got the second place ,a Got the third place .
        The results of the competition show that , They are only half right , And there is no parallel ranking , Output the ranking of students
 *
 * \ anaysis ： Yes abcd Go ahead separately 1234 enumeration , It can be realized by triple loop , final d use d=10-a-b-c;

/** \ question 2:3 A teacher predicted a math contest , Their predictions are as follows ：
         A said ：a Won the first place ,b Got the third place ;
         B said ：c Won the first place ,d Got the fourth place ;
         C said ：d Got the second place ,a Got the third place .
         The results of the competition show that , They are only half right , And there is no parallel ranking , Output the ranking of students 
 *
 * \ anaysis ： Yes abcd Go ahead separately 1234 enumeration , It can be realized by triple loop , final d use d=10-a-b-c;
 * \
 * \return
 *
 */

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int a,b,c,d;
    for(a=1;a<5;++a)
        for(b=1;b<5;++b)
            if(b!=a)
            for(c=1;c<5;++c)
            if(c!=a&&c!=b)
            {
                d=10-a-b-c;
                if(((a==1)+(b==3)==1)&&((c==1)+(d==4)==1)&&((d==2)+(a==3)==1))
                    printf("a  yes  %dth\nb  yes  %dth\nc  yes  %dth\nd  yes  %dth\n",a,b,c,d);
            }
    return 0;
}

problem 3：

PAT（ Class B ）2022 Summer exam

7-4 What day of the week is today

fraction 20

author Chen Yue

Company Zhejiang University

Han Meimei and Li Lei are arguing. What day of week is it today , Their dialogue is as follows ：

 Han Meimei ： Today is Friday .
 Li lei ： incorrect ！ Today is Saturday .
 Han Meimei ： But yesterday was Wednesday .
 Li lei ： impossible ！ Yesterday was Thursday .
 Han Meimei ： Tomorrow is Tuesday .
 Li lei ： Funny ？ Tomorrow is Monday .

Their mother told you , Each bear child only said one word right , Please judge , What day of week is today ？

Input format ：

The input contains 2 That's ok , Each line gives a statement of a bear child , The format is as follows ：

 yesterday   today   Tomorrow, 

The date here is 0 To 6 The integer of , Corresponding to Sunday to Saturday .

Output format ：

First, output the first line “ today ” What day is it , It is required to output its English name （ The comparison table is given after the example ）. Input to ensure that each bear child is only right in one sentence ,“ today ” The answer exists and is unique .

And then 2 Print out the day on which each bear child answers correctly —— yes yesterday It outputs yesterday, Or is it today To export today, Or is it Tomorrow, To export tomorrow.

sample input ：

3 5 2
4 6 1

sample output ：

Friday
today
yesterday

Be careful ： The English name of the date is as follows

 Sunday ：0 - Sunday
 Monday ：1 - Monday
 Tuesday ：2 - Tuesday
 Wednesday ：3 - Wednesday
 Thursday ：4 - Thursday
 Friday ：5 - Friday
 Saturday ：6 - Saturday

Code length limit

16 KB

The time limit

400 ms

Memory limit

64 MB

A total of three codes are written ： The first code I don't want to write for a lifetime .

coding first: Judge the number of days the second bear child tells the truth , It's really hard to judge , It can be said to be exhausted , The second program is easy to write , This can be directly ignored , Belong to is to give the fight out .

/**
     Judge the number of days the second bear child tells the truth , It's really hard to judge , It can be said to be exhausted ,
     The second program is easy to write 
*/

/**
#include <iostream>
#include <map>
#include <string>
using namespace std;

int main()
{
    int ch[2][3];
    for(int i=0;i<2;++i)
        for(int j=0;j<3;++j)
            scanf("%d",&ch[i][j]);
    int index=0;
    for(int i=-1;i<2;++i)
    {
        int yes=ch[0][i+1];
        if((((yes-ch[0][0])%7==i+1)+((yes-ch[0][1])%7==i)+((yes-ch[0][2])%7==i-1))==1
            &&(((yes-ch[1][0])%7==i+1)+((yes-ch[1][1])%7==i)+((yes-ch[1][2])%7==i-1)==1))
                index=i+1;
    }

    map<int,string> mp={
   {0,"Sunday"},{1,"Monday"},{2,"Tuesday"},{3,"Wednesday"},
                        {4,"Thursday"},{5,"Friday"},{6,"Saturday"}};
    int day=ch[0][index];
    if(index==0)
        cout << mp[(day+1)%7] << endl;
    else if(index==1)
        cout << mp[day] << endl;
    else cout << mp[(day-1)%7] << endl;

    map<int,string> str{
   {0,"yesterday"},{1,"today"},{2,"tomorrow"}};
    for(int i=0;i<3;++i)
        if(i==index)
        {
            cout << str[i] << endl;
        }

    for(int i=0;i<3;++i)
    {
        if(index<=i) // According to the truth of the first bear child , Introduce the truth that the second bear child said 
        {
            //if The statement can only be (5,0),(6,0)(6,1) To enter ;
            // because (index-i)%7 It can only be equal to 5,6,0((day-ch[1][i])%7 It can't be equal to 0）, therefore 0 Can be ruled out 
            if(day>ch[1][i] &&  (day-ch[1][i])%7==(index-i)%7)// If the first bear child told the truth yesterday and Friday 
                cout << str[i] << endl;                         // Or Saturday to judge the day when the second child is telling the truth 
            else if((day <= ch[1][i]) && day-ch[1][i]==index-i)
                cout << str[i] << endl;
        }

        else if(index>i && ((day-ch[1][i])%7==index-i))
            cout << str[i] << endl;
    }
    return 0;
}
*/

coding second : Because the title said that each bear child only said one sentence , We can find the subscript of the truth spoken by the second bear child in the same way as the first bear child speaks the truth

/**
     To write ： Because the title said that each bear child only said one sentence , We can say by looking for the first bear child 
     The same way to find out the subscript of the truth that the second bear child said 

*/
/**
#include <iostream>
#include <map>
#include <string>
using namespace std;

int main()
{
    int ch[2][3];
    for(int i=0;i<2;++i)
        for(int j=0;j<3;++j)
            scanf("%d",&ch[i][j]);
    int index=0;
    for(int i=-1;i<2;++i)
    {
        int yes=ch[0][i+1];
        if((((yes-ch[0][0])%7==i+1)+((yes-ch[0][1])%7==i)+((yes-ch[0][2])%7==i-1))==1
            &&(((yes-ch[1][0])%7==i+1)+((yes-ch[1][1])%7==i)+((yes-ch[1][2])%7==i-1)==1))
                index=i+1;  // Because the title said that each bear child only said one sentence , You can find out the truth that the first bear child said 
    }                       // The subscript 

    map<int,string> mp={
   {0,"Sunday"},{1,"Monday"},{2,"Tuesday"},{3,"Wednesday"},
                        {4,"Thursday"},{5,"Friday"},{6,"Saturday"}};
    int day=ch[0][index];
    if(index==0)
        cout << mp[(day+1)%7] << endl;
    else if(index==1)
        cout << mp[day] << endl;
    else cout << mp[(day-1)%7] << endl;

    map<int,string> str{
   {0,"yesterday"},{1,"today"},{2,"tomorrow"}};
    for(int i=0;i<3;++i)
        if(i==index)
        {
            cout << str[i] << endl;
        }

    for(int i=-1;i<2;++i)
    {
        int yes=ch[1][i+1];
        if((((yes-ch[0][0])%7==i+1)+((yes-ch[0][1])%7==i)+((yes-ch[0][2])%7==i-1))==1
            &&(((yes-ch[1][0])%7==i+1)+((yes-ch[1][1])%7==i)+((yes-ch[1][2])%7==i-1)==1))
                index=i+1;  // Because the title said that each bear child only said one sentence , We can say by looking for the first bear child 
    }                       // The same way to find out the subscript of the truth that the second bear child said 

    for(int i=0;i<3;++i)
        if(i==index)
        {
            cout << str[i] << endl;
        }
    return 0;
}
*/

coding thirt: It suddenly occurred to me that there was another solution , Let's assume today's date is （0~6）, Then use the words of two bear children to judge whether it is true ;

/**
     It suddenly occurred to me that there was another solution , Let's assume today's date is （0~6）, Then use the words of two bear children to judge whether it is true ;

*/

#include <iostream>
#include <map>
#include <string>
using namespace std;

int main()
{
    map<int,string> date={
   {0,"Sunday"},{1,"Monday"},{2,"Tuesday"},{3,"Wednesday"},
                        {4,"Thursday"},{5,"Friday"},{6,"Saturday"}};
    map<int,string> true_word{
   {0,"yesterday"},{1,"today"},{2,"tomorrow"}};
    int ch[2][3];
    for(int i=0;i<2;++i)
        for(int j=0;j<3;++j)
            scanf("%d",&ch[i][j]);
    int today=0;
    for(today =0;today<7;++today)
    {
        if(((((today-ch[0][0])%7==1)+((today-ch[0][1])==0)+((today-ch[0][2])%7==6))==1)&&
            (((today-ch[1][0])%7==1)+((today-ch[1][1])==0)+((today-ch[1][2])%7==6))==1)
                break;
    }

    cout << date[today] << endl;
    int index=0;
    for(int k=0;k<2;++k)
    {
        for(int i=0;i<3;++i)
            if(((today-ch[k][i])%7)==(1-i)%7)
                index=i;
        cout << true_word[index] << endl;
    }
}

prlblem 4：

1089 Werewolf killing - Simple version

fraction 20

author CHEN, Yue

Company Zhejiang University

The following text is taken from 《 brainwave · Fun math 》：“ Werewolf killing ” The game is divided into werewolves 、 The two camps of good people . In one game “ Werewolf killing ” In the game ,1 Player No. 1 said ：“2 Number one is a werewolf ”,2 Player No. 1 said ：“3 Number one is a good man ”,3 Player No. 1 said ：“4 Number one is a werewolf ”,4 Player No. 1 said ：“5 Number one is a good man ”,5 Player No. 1 said ：“4 Number one is a good man ”. Known 5 Of the players 2 Man plays the role of werewolf , Yes 2 People are not telling the truth , There are werewolves lying, but not all werewolves are lying . Which two players play the role of werewolf ？

This question is an upgraded version of this problem ： It is known that N Of the players 2 Man plays the role of werewolf , Yes 2 People are not telling the truth , There are werewolves lying, but not all werewolves are lying . Ask you to find out which players are playing the role of werewolves ？

Input format ：

Input gives a positive integer in the first line N（5≤N≤100）. And then N That's ok , The first i Line gives i What player No. 1 said （1≤i≤N）, That is, a player number , Use a plus sign for good people , A minus sign means a werewolf .

Output format ：

If there is any solution , Output in ascending order on a line 2 The number of a werewolf , Separated by spaces , There must be no extra space at the beginning and end of the line . If the solution is not unique , Then output the minimum sequence solution —— That is, for two sequences A=a[1],...,a[M] and B=b[1],...,b[M], If exist 0≤k<M bring a[i]=b[i] （i≤k）, And a[k+1]<b[k+1], It's called sequence A Less than sequence B. If there is no solution, the output is No Solution.

sample input 1：

5
-2
+3
-4
+5
+4

sample output 1：

1 4

sample input 2：

6
+6
+3
+1
-5
-2
+4

sample output 2（ The solution is not unique ）：

1 5

sample input 3：

5
-2
-3
-4
-5
-1

sample output 3：

No Solution

Code length limit

16 KB

The time limit

400 ms

Memory limit

64 MB

coding ： This code should be read by Liu Shen ：

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
    int n;
    cin >> n;
    vector<int> vec(n+1);
    for(int i=1;i<=n;++i)
        cin >> vec[i];
    for(int i=1;i<n;++i) // The first two floors for Cycle through werewolf numbers 
    {
        for(int j=i+1;j<=n;++j)
        {
            vector<int> goodman(n+1,1),lie; // lie Store the number of the liar 
            goodman[i]=goodman[j]=-1; // hypothesis i and j It's a werewolf 
            for(int k=0;k<=n;++k)
            {
                if(vec[k]*goodman[abs(vec[k])]<0) // If k What the number one said and what he said specify the true identity of the person 
                    lie.push_back(k);             // If not, the person told a lie 
            }
            if(lie.size()==2&&goodman[lie[0]]+goodman[lie[1]]==0) // If there are just two liars 
            {
                cout << i << ' ' << j << endl;  // And a person is a werewolf , One is a good man , Then I found the answer 
                return 0;
            }

        }
    }
    cout << "No Solution\n" ; // If none of them are found , There is no solution 
    return 0;
}