1、 求一个矩阵中的两个数，要求两个数不在一行一列，使得这两个数的乘积最大，输出最大积。直接暴力，5分钟做完
//矩阵
public class 题目1 {
public static void main(String[] args){
Scanner scanner = new Scanner (System.in);
int n=scanner.nextInt ();
int m = scanner.nextInt ();
int a[][] = new int[n][m];
int max = Integer.MIN_VALUE;
for (int i=0;i<a.length;i++){
for (int j=0;j<a[i].length;j++){
a[i][j] = scanner.nextInt ();
}
}
for (int i=0;i<a.length;i++){
for (int j=0;j<a[i].length;j++){
for (int x=0;x<a.length;x++){
for (int y = 0;y<a[x].length;y++){
if (x!=i&&y!=j) {
int sum = a[i][j] * a[x][y];
if (sum > max) {
max = sum;
}
}
}
}
}
}
System.out.println (max);
}
} 2、按给定排序方法给数组排序，这道题花了我一个半小时的时间，将归并、快排、希尔、插入、选择、堆排都调试了一遍，最后感觉归并最像给出的结果集，不知道是不是归并 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<a.length;i++){
a[i] =scanner.nextInt ();
}
mergeSort (a,0,a.length-1);
for (int i=0;i<a.length;i++){
System.out.print (a[i]+" " );
}
}
private static void mergeSort(int[] a, int left, int right) {
if (left>=right){
return;
}
int mid = (left+right)>>>1;
mergeSort (a,left,mid);
mergeSort (a,mid+1,right);
merge(a,left,mid,right);
}
private static void merge(int[] a, int left, int mid, int right) {
int s1 = left;
int s2 = mid+1;
int res[] = new int[right-left+1];
int i=0;
while (s1<=mid&&s2<=right){
if (a[s1]<=a[s2]){
res[i++] = a[s1++];
}else {
res[i++] = a[s2++];
}
}
while (s1<=mid){
res[i++] = a[s1++];
}
while (s2<=right){
res[i++] = a[s2++];
}
for (int j=0;j<res.length;j++){
a[j+left] = res[j];
}
} 3、给定一个数字，求转化为二进制后1的个数，没啥好说的五分钟 public static void main(String[] args){
Scanner scanner = new Scanner (System.in);
Long num = scanner.nextLong ();
String s = Long.toString (num, 2);
String s1 = s.replaceAll ("0", "");
System.out.println (s1.length ());
} 4、将字符串向左移位n个，比如hellowrold向左移位5个，变成worldhello，没啥好说的10分钟，一开始的时候没有考虑到n可能大于字符串长度的情况，耽误了点时间 public static void main(String[] args){
Scanner scanner = new Scanner (System.in);
String s1 = scanner.nextLine ();
int n = scanner.nextInt ();
if (n>=s1.length ()){
n = n%s1.length ();
}
char c1 [] = new char[n];
char c2 [] = new char[s1.length ()-n];
for (int i=0;i<s1.length ();i++){
if (i<n){
c1[i] = s1.charAt (i);
}else {
c2[i-n] = s1.charAt (i);
}
}
System.out.println (new String (c2)+new String (c1) );
} 
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