Bear and Prime 100（交互题）

Bear and Prime 100（CodeForces - 679A ）
https://codeforces.com/problemset/problem/679/A
题目：
This is an interactive problem. In the output section below you will see the information about flushing the output.

Bear Limak thinks of some hidden number — an integer from interval [2, 100]. Your task is to say if the hidden number is prime or composite.

Integer x > 1 is called prime if it has exactly two distinct divisors, 1 and x. If integer x > 1 is not prime, it’s called composite.

You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2, 100]. The system will answer “yes” if your integer is a divisor of the hidden number. Otherwise, the answer will be “no”.

For example, if the hidden number is 14 then the system will answer “yes” only if you print 2, 7 or 14.

When you are done asking queries, print “prime” or “composite” and terminate your program.

You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2, 100]. Also, you will get the Wrong Answer verdict if the printed answer isn’t correct.

You will get the Idleness Limit Exceeded verdict if you don’t print anything (but you should) or if you forget about flushing the output (more info below).

输入：
After each query you should read one string from the input. It will be “yes” if the printed integer is a divisor of the hidden number, and “no” otherwise.

输出：
Up to 20 times you can ask a query — print an integer from interval [2, 100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.

In any moment you can print the answer “prime” or “composite” (without the quotes). After that, flush the output and terminate your program.

To flush you can use (just after printing an integer and end-of-line):

fflush(stdout) in C++;
System.out.flush() in Java;
stdout.flush() in Python;
flush(output) in Pascal;
See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2, 100]. Of course, his/her solution won’t be able to read the hidden number from the input.

例子：

注意：
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.

The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn’t necessary to know the exact value of the hidden number. In this test, the hidden number is 30.

59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn’t forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).

题意：
这道题是到交互题，首先你需要确定一个数字，输出结果，后台会进行判定，然后根据这个数是否为隐藏数字的除数，然后输入“yes”或“no”，你再根据这个字符判定隐藏数字的性质，直到最高次数20次，或者判定出来性质，质数输出“prime”，合数输出“composite”，隐藏数字在【2,100】。
思路：
首先已知隐藏数字在【2,100】。对于 100 以内的数，如果为合数，其必为小于等于50的两个数相乘的结果，如：2*50=100，其是最大的一个结果50。可以对50以内的数进行打表，得到素数，在对素数进行 i * i 运算，得到一个数组，（i * i 运算是为了避免漏掉某些数字，例如：4 = 2 * 2，如果没有 4 ，在最后判定是只会判到一个2，会得到一个错误结论，如果有更好的方法可以在评论区留言分享），再进行以此判定。

t=19
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 4 9 25 49

数组共有19 个数字。

#include
using namespace std;
int a[100],t;
int main()
{
	for(int i=2;i<51;i++){
		int f=1;
		for(int j=2;j<=sqrt(i);j++){
			if(i%j==0){
				f=0;break;
			}
		}
		if(f) a[++t]=i;
	}
	for(int i=1;i<=t;i++) {
		if(a[i]*a[i]>100) break;
		else a[++t]=a[i]*a[i];
	}
	int n=20,f=2,i=1,k=2;
	while(n--&&f&&i<=t){
		k=a[i++];
		cout<<k<<endl;
		char s[5];
		cin>>s;
		if(s[0]=='y') f--;
	}
	if(!f) cout<<"composite"<<endl;
	else cout<<"prime"<<endl;
 }