文章目录
- A.重合次数(模拟)
- B.数数(数学)
A.重合次数(模拟)
Answer:494
这个题目比较坑人,
public class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
static int N = (int) (1e5 + 10);
public static void main(String[] args) throws Exception{
LocalTime st = LocalTime.of(6, 13, 22);
LocalTime ex = LocalTime.of(14, 36,21);
int ans = 0;
while(!st.equals(ex)) {
int min = st.getMinute();
int sed = st.getSecond();
if(min == sed) {
ans++;
}
st = st.plusSeconds(1);
}
System.out.println(ans);
}
}
//502 注意需要-8
接下来,我们用程序解释一下,为什么需要减去8,我这里给大家说的清楚一点,
#include <bits/stdc++.h>
using namespace std;
int main() {
double hs = 1.0 / 3600.0;
double ms = 1.0 / 60.0;
double ss = 1.0;
double sth = 6.0;
double stm = 13.0 + 22 * ms;
double sts = 22.0;
int ans = 0;
for (int i = 1; i<= 3600*2 ; i ++) {
cout << sth << ' ' << stm << ' ' << sts << endl;
if (abs(stm / 60.0 - sts / 60.0) < 1e-8)
ans ++;
sts += ss;
stm += ms;
if (sts == 60) sts = 0;
if (abs(stm - 60) < 1e-8) stm = 0, sth += 1;
if (sth == 14 && abs(stm-36) < 1e-8 && sts == 20.0) {
break;
}
}
cout << ans;
}
B.数数(数学)
Answer:25606
质因数分解定理解题即可
public class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
static int N = (int) (1e5 + 10);
static int cnt = 0;
static boolean isPrime[] = new boolean[N];
static long[] primes = new long[N];
static Map<Integer, Integer> map = new HashMap<>();
public static void main(String[] args) throws Exception {
int ans = 0;
for (int i = 2333333; i <= 23333333; i++) {
int ct = getDiv(i);
if(ct == 12) {
ans++;
}
}
System.out.println(ans);
}
private static int getDiv(int n) {
map.clear();
int num = 0;
int s = 0;
for(int i = 2; i <= n/i; i++) {
if(n % i == 0) {
s = 0;
while(n % i == 0) {
n = n / i;
s++;
}
num += s;
}
}
if(n > 1) {
num += 1;
}
return num;
}
}