科大讯飞，8.15 研发类笔试（二） 算法代码。
第一次全AC
1、定义一个n*m的数字矩阵，找出不同行不同列的两个数的乘积的最大值。
暴力，没想到能AC
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
long[][] arr = new long[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
arr[i][j] = scanner.nextLong();
}
}
long max = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
long one = arr[i][j];
for (int k = 0; k < n; k++) {
for (int z = 0; z < m; z++) {
if (k != i && z != j) {
long two = arr[k][z];
max = Math.max(one * two, max);
}
}
}
}
}
System.out.println(max);
} 2、插入排序，没什么好说的 3、求32位整数N的二进制1的个数。
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
long target = scanner.nextLong();
int count = 0;
while (target>0){
target&=(target-1);
count++;
}
System.out.println(count);
} 4、将一个字符串长度为m的字符串左移n位
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String m = scanner.nextLine();
int n = scanner.nextInt()% m.length();
System.out.println(m.substring(n,m.length()) + m.substring(0,n));
} 
