[C++知识库]PTA 7-1 猜数列游戏2-1 (10 分)

在猜数列游戏1的基础上，设置一个数组用来存放6个数列的第2到第4个数，在用户要求猜测新数列的情况下，循环显示各数列的前两个数，要求用户猜测第3个数。 6个数列3个组成的数组为：{1,2,3,3,4,7,2,5,12,3,6,10,4,9,16,5,12,22}

输入格式:

根据要求输入整数数值为字符Y或N

输出格式:

各提示信息与猜数列游戏1一致。

输入样例:

在这里给出一组输入。例如：

3
Y
7
Y
12
Y
10
Y
16
Y
22
Y
5
Y
3
N

输出样例:

在这里给出相应的输出。例如：

The first 2 value is 1,2.What is the next value?
Want to try another sequence?(Y/N)
The first 2 value is 3,4.What is the next value?
Want to try another sequence?(Y/N)
The first 2 value is 2,5.What is the next value?
Want to try another sequence?(Y/N)
The first 2 value is 3,6.What is the next value?
Want to try another sequence?(Y/N)
The first 2 value is 4,9.What is the next value?
Want to try another sequence?(Y/N)
The first 2 value is 5,12.What is the next value?
Want to try another sequence?(Y/N)
The first 2 value is 1,2.What is the next value?
Nice guess but not quite it.
Do you want to guess again?
Want to try another sequence?(Y/N)
num_of_rights:7 num_of_tries:8

代码实现（C++）

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
	int a1=3,a2=5,i=0,num,sum_t=0,sum_Y=0;
	int a[18]={1,2,3,3,4,7,2,5,12,3,6,10,4,9,16,5,12,22};
	int count=0; 
	char ch,ch2;
	bool go1=true,go2=true;
	while(1)
	{	
		if(i<1)
            cout << "The first 2 value is " << a[count] << ',' << a[count+1] << ".What is the next value?" << endl;
        i++;    
		if(i<=100&&i>=1)	
			cin >> num;
		while(1)
		{
			if(num!=a[count+2]&&i==1)
			{	
                printf("Nice guess but not quite it.\n");
				sum_t++;
			}
			if(num!=a[count+2]&&i==2)	
			{
                printf("Wrong a second time.\n");
				sum_t++;
			}
			if(num!=a[count+2]&&i==3)
			{
                printf("This is harder than it looks.\n");
				sum_t++;
			}
			if(num!=a[count+2]&&i==4)
			{
				sum_t++;
                printf("It must be getting pretty frustrating by now!\n");
                
			}
			if(num!=a[count+2])
			{
				printf("Do you want to guess again?\n");	
				cin >> ch2;
				if(ch2=='Y')
				{
					break;
				}
			} 
			if(num == a[count+2]||ch2=='N')
			{	
				i=0;
				if(num==a[count+2])
				{
					sum_Y++;
					sum_t++;
				}
                printf("Want to try another sequence?(Y/N)\n");
                if(count<15)
                {
                	count+=3;
				}
				else count=0;
				cin >> ch;
				if(ch == 'Y')
					break;
				else
				{
					printf("num_of_rights:%d num_of_tries:%d",sum_Y,sum_t);
					exit(0);
				} 
			}
		}
	}
	return 0;
}