一道古老的智力题——奇怪的问题

奇怪的问题
请回答下面10个问题，各题都恰有一个答案是正确的：
（1）第一个答案是B的问题是哪一个？
A、2
B、3
C、4
D、5
E、6
（2）恰好有两个连续问题的答案是一样的，它们是：
A、2，3
B、3，4
C、4，5
D、5，6
E、6，7
（3）本问题答案和哪一个问题的答案相同？
A、1
B、2
C、4
D、7
E、6
（4）答案是A的问题的个数是：
A、0
B、1
C、2
D、3
E、4
（5）本问题答案和哪一个问题的答案相同？
A、10
B、9
C、8
D、7
E、6
（6）答案是A的问题的个数和答案是什么的问题的个数相同？
A、B
B、C
C、D
D、E
E、以上都不是
（7）按照字母顺序，本问题的答案和下一个问题的答案相差几个字母？
A、4
B、3
C、2
D、1
E、0（注：A和B相差一个字母）
（8）答案是元音字母的问题的个数是：
A、2
B、3
C、4
D、5
E、6（注：A和E是元音字母）
（9）答案是辅音字母的问题的个数是：
A、一个质数
B、一个阶乘数
C、一个平方数
D、一个立方数
E、5的倍数
（10）本问题的答案是：
A、A
B、B
C、C
D、D
E、E

 

---

 

Solution as follow 

 

// good.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"

#include "math.h"
#include "stdio.h"

//题1  返回：0 正确  -1 1~5 错误
int NO1(int *Answer)
...{
    int i=1;
    while(i<=10)
    ...{
        if(Answer[i]==2)
        ...{
            switch(Answer[1])
            ...{
            case 1: 
                if(i==2)
                    return 0;
                else
                    return 1;
            case 2:
                if(i==3)
                    return 0;
                else
                    return 2;
            case 3:
                if(i==4)
                    return 0;
                else
                    return 3;
            case 4:
                if(i==5)
                    return 0;
                else
                    return 4;
            case 5:
                if(i==6)
                    return 0;
                else
                    return 5;
            default:
                return -1;
            }
        }
        i++;
    }
    return -1;
}

//题2  返回：0 正确  -1 1~5 错误
int NO2(int *Answer)
...{
    int i=1,n=0;
    while(i<10)
    ...{
        if(Answer[i]==Answer[i+1]) n++;
        i++;
    }
    if(n==1)
    ...{
        switch(Answer[2])
        ...{
        case 1:
            if(Answer[2]==Answer[3])
                return 0;
            else
                return 1;
        case 2:
            if(Answer[3]==Answer[4])
                return 0;
            else
                return 2;
        case 3:
            if(Answer[4]==Answer[5])
                return 0;
            else
                return 3;
        case 4:
            if(Answer[5]==Answer[6])
                return 0;
            else
                return 4;
        case 5:
            if(Answer[6]==Answer[7])
                return 0;
            else
                return 5;
        default:
            return -1;
        }
    }
    return -1;
}

//题3  返回：0 正确  -1 1~5 错误
int NO3(int *Answer)
...{
    switch(Answer[3])
    ...{
    case 1:
        if(Answer[3]==Answer[1])
            return 0;
        else
            return 1;
    case 2:
        if(Answer[3]==Answer[2])
            return 0;
        else
            return 2;
    case 3:
        if(Answer[3]==Answer[4])
            return 0;
        else
            return 3;
    case 4:
        if(Answer[3]==Answer[7])
            return 0;
        else
            return 4;
    case 5:
        if(Answer[3]==Answer[6])
            return 0;
        else
            return 5;
    default:
        return -1;
    }
    return -1;
}

//题4  返回：0 正确  -1 1~5 错误
int NO4(int *Answer)
...{
    int i=1,n=0;
    while(i<=10)
    ...{
        if(Answer[i]==1) n++;
        i++;
    }
    switch(Answer[4])
    ...{
    case 1:
        if(n==0)
            return 0;
        else
            return 1;
    case 2:
        if(n==1)
            return 0;
        else
            return 2;
    case 3:
        if(n==2)
            return 0;
        else
            return 3;
    case 4:
        if(n==3)
            return 0;
        else
            return 4;
    case 5:
        if(n==4)
            return 0;
        else
            return 5;
    default:
        return -1;
    }
    return -1;
}

//题5  返回：0 正确  -1 1~5 错误
int NO5(int *Answer)
...{
    switch(Answer[5])
    ...{
    case 1:
        if(Answer[5]==Answer[10])
            return 0;
        else
            return 1;
    case 2:
        if(Answer[5]==Answer[9])
            return 0;
        else
            return 2;
    case 3:
        if(Answer[5]==Answer[8])
            return 0;
        else
            return 3;
    case 4:
        if(Answer[5]==Answer[7])
            return 0;
        else
            return 4;
    case 5:
        if(Answer[5]==Answer[6])
            return 0;
        else
            return 5;
    default:
        return -1;
    }
    return -1;
}

//题6  返回：0 正确  -1 1~5 错误
int NO6(int *Answer)
...{
    int i=1,na=0,nb=0,nc=0,nd=0,ne=0;
    while(i<=10)
    ...{
        switch(Answer[i])
        ...{
        case 1:na++;break;
        case 2: nb++;break;
        case 3: nc++;break;
        case 4: nd++;break;
        case 5: ne++;break;
        default:return -1;break;
        }
        i++;
    }
    switch(Answer[6])
    ...{
    case 1:
        if(na==nb)
            return 0;
        else
            return 1;
    case 2:
        if(na==nc)
            return 0;
        else
            return 2;
    case 3:
        if(na==nd)
            return 0;
        else
            return 3;
    case 4:
        if(na==ne)
            return 0;
        else
            return 4;
    case 5:
        if((na!=nb)&&(na!=nc)&&(na!=nd)&&(na!=ne))
            return 0;
        else
            return 5;
    default:
        return -1;
    }
    return -1;
}

//题7  返回：0 正确  -1 1~5 错误
int NO7(int *Answer)
...{
    switch(Answer[7])
    ...{
    case 1:
        if(abs(Answer[7]-Answer[8])==4)
            return 0;
        else
            return 1;
    case 2:
        if(abs(Answer[7]-Answer[8])==3)
            return 0;
        else
            return 2;
    case 3:
        if(abs(Answer[7]-Answer[8])==2)
            return 0;
        else
            return 3;
    case 4:
        if(abs(Answer[7]-Answer[8])==1)
            return 0;
        else
            return 4;
    case 5:
        if(abs(Answer[7]-Answer[8])==0)
            return 0;
        else
            return 5;
    default:return -1;
    }
    return -1;
}

//题8  返回：0 正确  -1 1~5 错误
int NO8(int *Answer)
...{
    int i=1,n=0;
    while(i<=10)
    ...{
        if((Answer[i]==1)||(Answer[i]==5)) n++;
        i++;
    }
    switch(Answer[8])
    ...{
    case 1:
        if(n==2)
            return 0;
        else
            return 1;
    case 2:
        if(n==3)
            return 0;
        else
            return 2;
    case 3:
        if(n==4)
            return 0;
        else
            return 3;
    case 4:
        if(n==5)
            return 0;
        else
            return 4;
    case 5:
        if(n==6)
            return 0;
        else
            return 5;
    default:return -1;
    }
    return -1;
}
//题9  返回：0 正确  -1 1~5 错误
int NO9(int *Answer)
...{
    int i=1,n=0;
    while(i<=10)
    ...{
        if((Answer[i]==2)||(Answer[i]==3)||(Answer[i]==4)) n++;
        i++;
    }
    switch(Answer[9])
    ...{
    case 1:
        if((n==2)||(n==3)||(n==5)||(n==7))//质数(素数)
            return 0;
        else
            return 1;
    case 2:
        if((n==1)||(n==2)||(n==6))//阶乘数
            return 0;
        else
            return 2;
    case 3:
        if((n==1)||(n==4)||(n==9))//平方数
            return 0;
        else
            return 3;
    case 4:
        if((n==1)||(n==8))//立方数
            return 0;
        else
            return 4;
    case 5:
        if((n==5)||(n==10))//5的倍数
            return 0;
        else
            return 5;
    default:return -1;
    }
    return -1;
}


int _tmain(int argc, _TCHAR* argv[])
...{
    int i;
    int Answer[11];
    char Answerc[5] = ...{'a','b','c','d','e'};
    for (Answer[1]=1;Answer[1]<=5;Answer[1]++)
        for (Answer[2]=1;Answer[2]<=5;Answer[2]++)
            for (Answer[3]=1;Answer[3]<=5;Answer[3]++)
                for (Answer[4]=1;Answer[4]<=5;Answer[4]++)
                    for (Answer[5]=1;Answer[5]<=5;Answer[5]++)
                        for (Answer[6]=1;Answer[6]<=5;Answer[6]++)
                            for (Answer[7]=1;Answer[7]<=5;Answer[7]++)
                                for (Answer[8]=1;Answer[8]<=5;Answer[8]++)
                                    for (Answer[9]=1;Answer[9]<=5;Answer[9]++)
                                        for (Answer[10]=1;Answer[10]<=5;Answer[10]++)
                                        ...{
                                            if((NO1(Answer)==0)&&(NO2(Answer)==0)&&(NO3(Answer)==0)&&(NO4(Answer)==0)&&(NO5(Answer)==0)&&(NO6(Answer)==0)&&(NO7(Answer)==0)&&(NO8(Answer)==0)&&(NO9(Answer)==0))
                                            ...{
                                                for(i = 1; i < 11; i++)...{
                                                    printf(" %c ",Answerc[Answer[i] - 1]);
                                                }
                                            }
                                        }
                                        return 0;
}