A: 按照题意来。。
import java.util.Scanner;
/**
* Created by dezhonger on 2019/4/13
*/
public class A {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int a = scanner.nextInt();
int b = scanner.nextInt();
long r = 0;
if (a > b) {
r += a;
a--;
} else {
r+= b;
b--;
}
r += Math.max(a, b);
System.out.println(r);
}
}
B: 按照题意来。。
import java.util.Scanner;
/**
* Created by dezhonger on 2019/4/13
*/
public class B {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = scanner.nextInt();
int max = Integer.MIN_VALUE;
int r = 0;
for (int i = 0; i < n ;i++) {
max = Math.max(max, a[i]);
if (max == a[i]) r++;
}
System.out.println(r);
}
}
C:
题意:
要相邻的两个数不同,问最少要改变多少个数
分析:要么是01010101..... 要么是101010101....
import java.util.Scanner;
/**
* Created by dezhonger on 2019/4/13
*/
public class C {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String s = scanner.nextLine();
char[] c = new char[]{'0', '1'};
int cc = 0;
int count1 = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != c[cc]) {
count1++;
}
cc = 1 - cc;
}
int count2 = 0;
cc = 1;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != c[cc]) {
count2++;
}
cc = 1 - cc;
}
System.out.println(Math.min(count1, count2));
}
}
D:可以选择把一个区间的0变为1,最多选k个区间,求最长连续1的个数
和leetcode1004相似
import java.util.Scanner;
/**
* Created by dezhonger on 2019/4/13
*/
public class D {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int k = scanner.nextInt();
String s = scanner.next();
int zero = 0;
int left = 0;
int res = 0;
for (int right = 0; right < n; ++right) {
if (s.charAt(right) == '0') {
zero++;
for (int j = right + 1; j < n && s.charAt(j) == '0'; j++) {
right = j;
}
while (zero > k) {
if (s.charAt(left) == '0') {
--zero;
for (int j = left; j < n && s.charAt(j) == '0'; j++) {
left = j;
}
}
left++;
}
}
res = Math.max(res, right - left + 1);
}
System.out.println(res);
}
}