public class Test {
private int HASH_SIZE=(1 HASH_LOG);
private int HASH_MASK=(HASH_SIZE - 1);
public Long flz_hash(Long a){
return index HASH_MASK;
}
public static void main(String[] args) {
Test test = new Test();
Long index = test.flz_hash(a);
System.out.println(index);
完整的直译为java方法是这样的,但是这样写dataLen的值必须小于等于数组长度,不然会出现下标越界的异常
public?int?accumulativeVerification(int[]?data,?int?dataLen)?{
int?sum?=?0;
for?(int?i?=?0;?i?dataLen;?i++)?{
sum?+=?data[i];
return?sum;
如果是意译(计算int数组的所有数之和),那么java可以这样写
public?int?accumulativeVerification(int[]?data)?{
for?(int?i?=?0;?i?data.length;?i++)?{
import java.util.Arrays;
import java.util.Scanner;
public class A {
static long[] a=new long[maxn],t=new long[maxn];
static long ans=0;
static void mymerge(int x,int m,int y) {
int tx=x,tm=m+1;
int k=0;
while(tx=m tm=y) {
if(a[tx]=a[tm])
t[k++]=a[tx++];
else {
ans+=m-tx+1;
a[k++]=a[tm++];
while(tx=m) t[k++]=a[tx++];
while(tm=y) t[k++]=a[tm++];
for(int i=0;ik;i++)
a[x++]=t[i];
static void mergesort(int x,int y) {
if(x==y)return ;
int xx=x,yy=y;
mergesort(xx,mid);
mergesort(mid+1,yy);
mymerge(xx,mid,yy);
//System.out.println(xx+" "+mid+" "+yy);
public static void main(String[] arg) {
int n,k;
Scanner input=new Scanner(System.in);
while(input.hasNext()) {
n=input.nextInt();
k=input.nextInt();
Arrays.fill(a, 0);
Arrays.fill(t, 0);
for(int i=0;in;i++) {
a[i]=input.nextLong();
ans=0;
mergesort(0, n-1);
if(kans)
System.out.println("0");
else
System.out.println((ans-k)+"");
package dis;
import java.io.*;
import java.math.*;
public class dis
{
public static void main(String [] s) throws Exception
Float a,b,c;
double x1;
System.out.println("请输入a, b, c的值");
BufferedReader bReader=new BufferedReader(new InputStreamReader(System.in));
a=Float.parseFloat(bReader.readLine());
b=Float.parseFloat(bReader.readLine());
c=Float.parseFloat(bReader.readLine());
System.out.println("x1 = "+ x1);
return;
unsigned short crc(unsigned char [] crc_pointer,unsigned short data_length)
unsigned short k,k0,bit_flag;
unsigned short int_crc=0xffff;
for (k=0;kdata_length;k++)
int_crc^=(crc_pointer)[k];
bit_flag=int_crc0x0001;
int_crc=1;
if (bit_flag==1)
int_crc^=0xa001;
return(int_crc);
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