Codeforces 680C

A. Bear and Prime 100

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Input

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.

Output

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.

Examples

input

Copy

yes
no
yes

output

Copy

2
80
5
composite

input

Copy

no
yes
no
no
no

output

Copy

58
59
78
78
2
prime

Note

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：通过不超过20次的询问输出，判断该数是否为素数

分析：每一个合数都可以分解为若干个素因子的乘积，2到100共有25个素因子，其中小于50的素因子有15个。大于50的合数一定会被分解为小于50的素因子的乘积，所以判断小于50的素因子即可。在判断时如果有两个及以上的素因子则一定是合数，如果只有一个素因子有可能是这个因子的幂，比如2^2,2^3...再判断i*i小于50的数即可（剩下的幂都含有这些因子）

代码：

#include<bits/stdc++.h>
using namespace std;
int isprime(int x)
{
    for(int i = 2; i <= sqrt(x); i++)
    {
        if(x % i == 0)
            return 0;
    }
    return 1;
}
int main()
{
    int cnt = 0;
    string s;
    for(int i = 2; i <= 50 && cnt < 2; i++)
    {
        if(isprime(i))
        {
            cout<<i<<endl;
            fflush(stdout);
            cin>>s;
            fflush(stdin);
            if(s == "yes")
                cnt++;
            if(i * i <= 50)
            {
                cout<<i * i<<endl;
                fflush(stdout);
                cin>>s;
                fflush(stdin);
                if(s == "yes")
                    cnt++;
            }
        }
    }
    if(cnt >= 2)
        printf("composite\n");
    else
        printf("prime\n");
    return 0;
}