小编use*_*112的帖子

以下是HackerEarth挑战赛中的一个问题 -

Roy有一个大小为NxN的矩阵.行和列的编号从1到N.第i行的第j列包含整数除法i/j.

换句话说,Matrix [i] [j] = int(i/j)其中1≤i,j≤N.

Your task is to find sum of this matrix i.e.

sum = 0
for i=1 to N-1
    for j=1 to N-1
        sum += Matrix[i][j]
Constraints:
1 ? T ? 10
1 ? N ? 1000000

这是我解决这个问题的方法

#include <cstdio>
#include <cassert>
using namespace std;
#define MAXT 10
#define MAXN 1000000
long long solve(long long N){
    long long ans = 0;
    for(int i=1;i<N-1;i++)
    {
    for(int j=1; j<N-1 ; j++)
    {
        int temp = N*i/j; …

编译时出错:

raja@raja-desktop:~/socket$ g++ Socket.cpp
Socket.cpp: In member function ‘int Socket::recv(std::string&) const’:
Socket.cpp:135: error: ‘cout’ is not a member of ‘std’

Socket.cpp的来源:

// Implementation of the Socket class.


#include "Socket.h"
#include "string.h"
#include <string.h>
#include <errno.h>
#include <fcntl.h>



Socket::Socket() :
  m_sock ( -1 )
{

  memset ( &m_addr,
    0,
    sizeof ( m_addr ) );

}

Socket::~Socket()
{
  if ( is_valid() )
    ::close ( m_sock );
}

bool Socket::create()
{
  m_sock = socket ( AF_INET,
      SOCK_STREAM,
      0 );

  if ( ! is_valid() …