枚举
视频讲解
🎥 视频讲解
枚举
枚举是指用循环的方式遍历问题中的全部情况，然后一一验证是否符合要求。
例题1
代码实现
参考实现
#include <bits/stdc++.h>
using namespace std;
bool check(int x) {
string s = " " + to_string(x);
int l = 1, r = s.size() - 1;
while (l < r) {
if (s[l] != s[r]) return false;
l++;
r--;
}
return true;
}
int main() {
int n, ans = 0;
cin >> n;
for (int i = 1; i <= n; i++) {
if (check(i)) ans++;
}
cout << ans;
return 0;
}
import java.util.Scanner;
public class Main {
static boolean check(int x) {
String s = " " + Integer.toString(x);
int l = 1, r = s.length() - 1;
while (l < r) {
if (s.charAt(l) != s.charAt(r)) {
return false;
}
l++;
r--;
}
return true;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int ans = 0;
for (int i = 1; i <= n; i++) {
if (check(i)) ans++;
}
System.out.println(ans);
}
}
def check(x: int) -> bool:
s = " " + str(x)
l, r = 1, len(s) - 1
while l < r:
if s[l] != s[r]:
return False
l += 1
r -= 1
return True
n = int(input())
ans = 0
for i in range(1, n + 1):
if check(i):
ans += 1
print(ans)
例题2
代码实现
参考实现
#include <bits/stdc++.h>
using namespace std;
bool check(int x) {
int cnt = 1; // 从个位开始，第 1 位
while (x) {
int d = x % 10;
if (cnt % 2 == 0) { // 偶数位
if (d % 2 != 0) return false;
} else { // 奇数位
if (d % 2 == 0) return false;
}
x /= 10;
cnt++;
}
return true;
}
int main() {
int n, ans = 0;
cin >> n;
for (int i = 1; i <= n; i++) {
if (check(i)) ans++;
}
cout << ans;
return 0;
}
import java.util.Scanner;
public class Main {
static boolean check(int x) {
int cnt = 1; // 从个位开始
while (x > 0) {
int d = x % 10;
if (cnt % 2 == 0) { // 偶数位
if (d % 2 != 0) return false;
} else { // 奇数位
if (d % 2 == 0) return false;
}
x /= 10;
cnt++;
}
return true;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int ans = 0;
for (int i = 1; i <= n; i++) {
if (check(i)) ans++;
}
System.out.println(ans);
}
}
def check(x: int) -> bool:
cnt = 1 # 从个位开始
while x > 0:
d = x % 10
if cnt % 2 == 0: # 偶数位
if d % 2 != 0:
return False
else: # 奇数位
if d % 2 == 0:
return False
x //= 10
cnt += 1
return True
n = int(input())
ans = 0
for i in range(1, n + 1):
if check(i):
ans += 1
print(ans)
例题3
代码实现
参考实现
#include <bits/stdc++.h>
using namespace std;
int n, k, t;
int a[10010];
int main() {
cin >> t;
while (t--) {
cin >> n >> k;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
int ans = 0x3f3f3f3f;
for (int color = 1; color <= 60; color++) {
int cnt = 0;
for (int j = 1; j <= n; j++) {
if (a[j] != color) {
cnt++;
j += k - 1;
}
}
ans = min(ans, cnt);
}
cout << ans << '\n';
}
return 0;
}
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int k = sc.nextInt();
int[] a = new int[n + 1];
for (int i = 1; i <= n; i++) {
a[i] = sc.nextInt();
}
int ans = 0x3f3f3f3f;
for (int color = 1; color <= 60; color++) {
int cnt = 0;
for (int j = 1; j <= n; j++) {
if (a[j] != color) {
cnt++;
j += k - 1;
}
}
ans = Math.min(ans, cnt);
}
System.out.println(ans);
}
}
}
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
a = [0] + list(map(int, input().split()))
ans = 10**9
for color in range(1, 61):
cnt = 0
j = 1
while j <= n:
if a[j] != color:
cnt += 1
j += k
else:
j += 1
ans = min(ans, cnt)
print(ans)