爱奇艺GG了
看到好多人2.6以上，0.7的表示GG。题确实简单，就是半天都优化不动。。。Java岗的，有大佬帮忙看一下怎么优化吗？
第一题：循环数比较大小，永远只有0.3的通过率
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long x1 = sc.nextLong();
long k1 = sc.nextLong();
long x2 = sc.nextLong();
long k2 = sc.nextLong();
if(x1 >= 1 && x1 <= 1000000000 &&
x2 >= 1 && x2 <= 1000000000 &&
k1 >= 1 && k1 <= 50 &&
k2 >= 1 && k2 <= 50) {
if(repeat(x1, k1) < repeat(x2, k2)) {
System.out.println("Less");
} else if(repeat(x1, k1) > repeat(x2, k2)) {
System.out.println("Greater");
} else if(repeat(x1, k1) == repeat(x2, k2)){
System.out.println("Equal");
}
}
}
public static long repeat(long x, long k) {
String str = String.valueOf(x);
StringBuffer sb = new StringBuffer();
for(int i = 0; i < k; i++) {
sb.append(str);
}
str = sb.toString();
long result = Integer.parseInt(str);
return result;
}
}
第三题：计算有序对数量
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
long m = sc.nextLong();
if(n >= 1 && n <= 100000 && m >= 1 && m <= 100000) {
System.out.println(SSR(n, m));
}
}
public static long SSR(long n, long m) {
long count = 0;
for(long A = 1; A <= n; A++) {
for(long B = 1; B <= m; B++) {
if((Math.sqrt(A) + Math.sqrt(B)) * (Math.sqrt(A) + Math.sqrt(B)) % 1 == 0) {
count++;
}
}
}
return count;
}
}
应该都不是超时，牛客的系统超时好像要提示。
算法太卷啦
