C. 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

yes
no
yes

Output

2
80
5
composite

Input

no
yes
no
no
no

Output

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).

題意：interesting的一道題目，讓你寫一個程式，可以辨別出一個數是否是素數，相當於猜數字，你自己在心中決定一個數然後讓你的程式來猜，如果程式輸出的數是你想的數的因子，輸入yes，否則輸入no，讓你寫這樣一個程式。

思路：因為資料範圍只有100，我們可以把100裡的素數表打出來，我們可以發現100以內的非素數都是2,3,5,7中的一個或者多個的倍數，這樣我們就可以列舉這四個數，然後控制電腦輸出就好了。

ac程式碼：

/* ***********************************************
Author       : AnICoo1
Created Time : 2016-08-21-20.51 Sunday
File Name    : D:\MyCode\2016-8月\2016-8-20.cpp
LANGUAGE     : C++
Copyright  2016 clh All Rights Reserved
************************************************ */
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define LL long long
#define ll __int64
#define mem(x,y) memset(x,(y),sizeof(x))
#define PI acos(-1)
#define gn (sqrt(5.0)+1)/2
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
const ll INF=1e9+10;
const int MAXN=1e6+10;
//head
int prime[110],vis[110],cnt;

void getprime()
{
    cnt=0;mem(vis,0);vis[1]=1;
    for(int i=2;i<=100;i++)
    {
        if(!vis[i])
        {
            prime[cnt++]=i;
            for(int j=i*2;j<=100;j+=i)
                vis[j]=1;
        }
    }
}

int main()
{
    getprime();char ch[5];
    for(int i=0;i<4;i++)
    {
        printf("%d\n",prime[i]);
        fflush(stdout);
        scanf("%s",ch);
        if(strcmp(ch,"yes")==0)
        {
            for(int j=i;j<cnt;j++)
            {
                if(prime[i]*prime[j]>100) break;
                printf("%d\n",prime[i]*prime[j]);
                fflush(stdout);
                scanf("%s",ch);
                if(strcmp(ch,"yes")==0)
                {
                    printf("composite\n");
                    fflush(stdout);
                    return 0;
                }
            }
        }
    }
    printf("prime\n");
    fflush(stdout);
    return 0;
}