Programming Question-6

最新推荐文章于 2021-09-01 21:01:34 发布

Gary_D_Shi

最新推荐文章于 2021-09-01 21:01:34 发布

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分类专栏： Coursera 算法课程文章标签： insert algorithm file fp integer numbers

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本文介绍了解决两个经典算法问题的方法：一是从文件中找出特定区间内所有能由不同整数对相加得到的目标值数量；二是通过不断接收整数流并实时计算中位数来维护一系列整数的中位数。

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1.Two-Sum Problem

The goal of this problem is to implement a variant of the 2-SUM algorithm (covered in the Week 5 lecture on hash table applications)

The file contains 500,000 positive integers (there might be some repetitions!).This is your array of integers, with the ith row of the file specifying the ithentry of the array.

Your task is to compute the number of target values t in the interval [2500,4000] (inclusive) such that there are distinct numbers x,y in the input file that satisfy x+y=t. (NOTE: ensuring distinctness requires a one-line addition to the algorithm from lecture.)

Write your numeric answer (an integer between 0 and 1501) in the space provided.

#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>
const char INFILE[] = "HashInt.txt";
#define MAX	10000000

void fetch_hash(long *hash)	{
  FILE *fp = fopen(INFILE,"r");
  char line[10];
  long i;
  for (i=0; i<MAX; i++)
	hash[i] = 0;
  while (fgets(line,10,fp))
	hash[atoi(line)] ++;
}
short two_sum(long *hash, long t)	{
  long x;
  for (x = 1; x < (t-1)/2; x++)	
	if ((hash[x])&&(hash[t-x]))
	  return 1;
  return 0;
}
 
void main()
{
  long *hash = calloc(MAX,sizeof(long));
  fetch_hash(hash);
  short i,count = 0;
  for (i = 2500; i <= 4000; i++)
	if (two_sum(hash,i))
	  count++;
  printf("count = %d\n",count);
}

2.Median Problem

The goal of this problem is to implement the "Median Maintenance" algorithm (covered in the Week 5 lecture on heap applications). The text file contains a list of the integers from 1 to 10000 in unsorted order; you should treat this as a stream of numbers, arriving one by one. Letting xi denote the ith number of the file, the kth median mk is defined as the median of the numbers x1,…,xk. (So, if k is odd, then mk is ((k+1)/2)th smallest number among x1,…,xk; if k is even, then mk is the (k/2)th smallest number among x1,…,xk.)

In the box below you should type the sum of these 10000 medians, modulo 10000 (i.e., only the last 4 digits). That is, you should compute (m1+m2+m3+⋯+m10000)mod10000.

#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>
#include"heap.h"
const char INFILE[] = "Median.txt";
#define MAX	10000

short RETURN_MEDIAN(short *A1,short *A2,short x)	{
  if (x<A2[1])
	MAX_HEAP_INSERT(A1,x);
  else
	MIN_HEAP_INSERT(A2,x);
  if (A1[0]-A2[0]>1)	
	MIN_HEAP_INSERT(A2,EXTRACT_MAX(A1));
  if (A2[0]>A1[0])	
	MAX_HEAP_INSERT(A1,EXTRACT_MIN(A2));
  return A1[1];
}
void print(short *A)	{
  short i;
  for (i=1; i<=A[0]; i++)
	printf("%d ",A[i]);
}

void main()	{
  short *A1,*A2,m=0;
  A1 = calloc(1+MAX,sizeof(short));
  A2 = calloc(1+MAX,sizeof(short));
  A1[0] = 1; A1[1] = 0;
  A2[0] = 1; A2[1] = 20000;
  char line[10];
  FILE *fp = fopen(INFILE,"r");
  while (fgets(line,10,fp))	
	m = (m + RETURN_MEDIAN(A1,A2,atoi(line))) % 10000;
  printf("%d\n",m);
}

where "heap.h" contains many heap functions.

#ifndef _HEAP_H_
#define _HEAP_H_
//	MAX_HEAP
void MAX_HEAPIFY(short *A,short i)	{
	short left=2*i,right=2*i+1,larger,tmp;
	if ((left<=A[0])&&(A[left]>A[i]))
	  larger = left;
	else
	  larger = i;
	if ((right<=A[0])&&(A[right]>A[larger]))
	  larger = right;
	if (larger != i)	{
	  tmp = A[larger]; A[larger] = A[i]; A[i] = tmp;
	  MAX_HEAPIFY(A,larger);
	}
}
short EXTRACT_MAX(short *A)	{
  short max = A[1];
  A[1] = A[A[0]--];
  MAX_HEAPIFY(A,1);
  return max;
}
void HEAP_INCREASE_KEY(short *A,short i,short key)	{
  short tmp;
  A[i] = key;
  while ((i>1)&&(A[i/2]<A[i]))	{
	tmp = A[i]; A[i] = A[i/2]; A[i/2] = tmp;
	i = i/2;
  }
}
void MAX_HEAP_INSERT(short *A,short key)	{
  A[0]++;
  HEAP_INCREASE_KEY(A,A[0],key);
}
//	MIN_HEAP
void MIN_HEAPIFY(short *A,short i)	{
	short left=2*i,right=2*i+1,smaller,tmp;
	if ((left<=A[0])&&(A[left]<A[i]))
	  smaller = left;
	else
	  smaller = i;
	if ((right<=A[0])&&(A[right]<A[smaller]))
	  smaller = right;
	if (smaller != i)	{
	  tmp = A[smaller]; A[smaller] = A[i]; A[i] = tmp;
	  MIN_HEAPIFY(A,smaller);
	}
}
short EXTRACT_MIN(short *A)	{
  short min = A[1];
  A[1] = A[A[0]--];
  MIN_HEAPIFY(A,1);
  return min;
}
void HEAP_DECREASE_KEY(short *A,short i,short key)	{
  short tmp;
  A[i] = key;
  while ((i>1)&&(A[i/2]>A[i]))	{
	tmp = A[i]; A[i] = A[i/2]; A[i/2] = tmp;
	i = i/2;
  }
}
void MIN_HEAP_INSERT(short *A,short key)	{
  A[0]++;
  HEAP_DECREASE_KEY(A,A[0],key);
}
#endif