非常抱歉，我无法写下 +、-、*、/ 之类的敏感的符号，所以只能自己手写了

template <class _Tp = const unsigned long long int>
class P { _Tp *p { nullptr }; };
signed WinMain(int argv, char argc[]) {}
#pragma GCC system_header
#include <cstdio>
#include <ctype.h>
typedef unsigned long long ULL;
int add(int a, int b) {
	if (!a) return b;
	return add((a & b) << 1, a ^ b);
}
ULL quick_mul(int a, int b) {
	ULL res = 0;
	for (ULL t = a; b; b >>= 1, t <<= 1)
		if (b & 1) res = add(res, t);
	return res;
}
ULL square(int a) {
	return quick_mul(a, a);
}
int read() {
	char ch;
	int res = 0;
	while (!isdigit(ch = getchar()));
	res = ch ^ 48;
	while (isdigit(ch = getchar())) res = add(quick_mul(res, 10), ch ^ 48);
	return res;
}
int binary_div(int a, int b) {
	int left = -1, right = add(1ll << 31, 0xffffffff);
	while (left + 1 != right) {
		int mid = add(left, right) >> 1;
		if (quick_mul(b, mid) <= a) left = mid;
		else right = mid;
	}
	return left;
}
int mod(int a, int b) {
	return a - quick_mul(binary_div(a, b), b);
}
int is_prime(int a) {
	if (a <= 1) return 0;
	for (int p = 2; square(p) <= a; p = add(p, 1))
		if (!mod(a, p)) return 0;
	return 1;
}
void print(int a) {
	if (!a) return;
	print(binary_div(a, 10));
	putchar(mod(a, 10) ^ 48);
}
signed main(int argv, char argc[]) {
	int a = read(), b = read();
	if (is_prime(a) & is_prime(b)) puts("yes");
	else if (is_prime(a) ^ is_prime(b)) puts("I can not do it");
	else puts("no");
	return 0;
}

什么难题，我看是水题

设个函数 isprime 来判断是否为质数即可：
记得用 cmath 库。

#include<iostream>
#include<cmath>
#define QwQ return 0;
using namespace std;
int n,m;
bool isprime(int x){
	if(x<2)return false;
	for(int i=2;i<=sqrt(x);i++){
		if(x%i==0)return false;
	}
	return true;
}
int main(){
	cin>>n>>m;
	switch(isprime(n)+isprime(m)){
		case 2:
			cout<<"yes";
			break;
		case 0:
			cout<<"no";
			break;
		case 1:
			cout<<"I can not do it";
			break;
	}
	QwQ
}

#include<iostream>
using namespace std;
bool IsPrime(int n){
	if(n==1)return false;
	for(int i=2;i*i<=n;i++)if(n%i==0)return false;
	return true;
}
int n,m;
signed main(){
	cin>>n>>m;
	if(IsPrime(n)&&IsPrime(m))cout<<"yes";
	else if(!IsPrime(n)&&!IsPrime(m))cout<<"no";
	else cout<<"I can not do it";
}

秀一波质数筛。

#include<bits/stdc++.h>
#define int long long
#define INF 0x3f3f3f
using namespace std;
int a[114514],n,m;
signed main(){
	for(int i=2;i<=1000;i++)
		a[i]=1;
	for(int i=2;i<=1000;i++){
		if(a[i]){
			for(int j=2;i*j<=1000;j++)
				a[i*j]=0;
		}
	}//质数筛
	cin>>n>>m;
	if(a[n]&&a[m])cout<<"yes";
	else if(!(a[n])&&!(a[m]))cout<<"no";
	else cout<<"I can not do it";
	return 0;
}

这题直接体现出了定义函数的重要性。

#include<bits/stdc++.h>
using namespace std;
bool prime(int a)
{
	if(a==1)
	{
		return 0;
	}
	for(int i=2;i*i<=a;i++)
	{
		if(a%i==0)
		{
			return 0;
		}
	}
	return 1;
}
int main()
{
	int a,b;
	cin>>a>>b;
	if(prime(a)&&prime(b))
	{
		cout<<"yes";
	}
	else if(!prime(a)&&!prime(b))
	{
		cout<<"no";
	}
	else
	{
		cout<<"I can not do it";
	}
	return 0;
}

C++ :

#include<bits/stdc++.h>
using namespace std;
int n,m,d,q;

int main()
{
cin>>n>>m;
for(int i=2;i<=sqrt(n);i++)
	{
	if(n%i==0)
		{
		q++;
		break;
		}
	}
if(n==1|n==0)
	q++;

for(int i=2;i<=sqrt(m);i++)
	{
	if(m%i==0)
		{
		q++;
		break;
		}
	}
if(m==1||m==0)
	q++;

if(q==0)
	cout<<"yes";
if(q==1)
	cout<<"I can not do it";
if(q==2)
	cout<<"no";
}