- 题解
徐木繁，张翌宸快抄
-
马浩元 (123456)
LV 6
@
2026-5-9 21:03:11
f #include using namespace std; int main(){ int n; cin >> n; int count = 0; for(int i = 2; i < n; i++){ if(n % i == 0) { count++; break; } } if(count == 0) cout << "Yes"; else cout << "No"; return 0; }
g.
#include using namespace std;
int main() {
int N, m;
cin >> N >> m;
int count = 0;
int num;
for (int i = 0; i < N; i++) {
cin >> num;
if (num == m) {
count++;
}
}
cout << count << endl;
return 0;
}
h.
#include using namespace std;
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) {
cout << i << endl;
}
return 0;
}
3 条评论
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f. #include using namespace std; int main(){ int n; cin >> n; int count = 0; for(int i = 2; i < n; i++){ if(n % i == 0) { count++; break; } } if(count == 0) cout << "Yes"; else cout << "No"; return 0; }
g.
#include using namespace std;
int main() { int N, m; cin >> N >> m; int count = 0; int num; for (int i = 0; i < N; i++) { cin >> num; if (num == m) { count++; } } cout << count << endl; return 0; }
h.
#include using namespace std;
int main() {
int n; cin >> n; for (int i = 1; i <= n; ++i) { cout << i << endl; }
return 0; }
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