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c++ code：（5）sort

news 来源：原创 2024/5/10 5:39:44

/*
程序填空题，自己编写排序函数 mysort，使得其能够对任意类型的数组排序


输入：
多组数据。每组数据以整数 n开头(n<10)，然后是n个整数

输出：
对每组数据，输出三行。
第一行是整数从小倒大排序的结果
第二行是按个位数从小到大排序的结果(个位数相同，则小的排在前面）
第三行还是整数从小倒大排序的结果




样例输入：

5 21 3 76 48 445
6 73 29 45 8737 2 1

样例输出
3,21,48,76,445,
21,3,445,76,48,
3,21,48,76,445,
1,2,29,45,73,8737,
1,2,73,45,8737,29,
1,2,29,45,73,8737,
*/



#include <iostream>
using namespace std;
struct A {
	int nouse1;
	int nouse2;
	int n;
};
//your code starts here
//bubble sort
void mysort(void * _array, int memnum, int typenum, int (*compare)(const void *,const void *))
{
	char * poi = (char *)_array;
	for (int i = memnum-1; i >=0; i--)
	{
		for (int j = 0; j<i; j++)
		{
			char * p1 = (char *)_array + j*typenum;
			char * p2 = (char *)_array + j*typenum+ typenum;
			int tmp = compare(p1,p2);
			if (tmp>0)
			{
				for (int k = 0; k < typenum; ++k)
				{
					char element= p1[k];
					p1[k] = p2[k];
					p2[k] = element;
				}
			}

		}
	}
}
//your code ends here		

int MyCompare1(const void * e1, const void * e2)
{
	int * p1 = (int *)e1;
	int * p2 = (int *)e2;
	return *p1 - *p2;
}
int MyCompare2(const void * e1, const void * e2)
{
	int * p1 = (int *)e1;
	int * p2 = (int *)e2;
	if ((*p1 % 10) - (*p2 % 10))
		return (*p1 % 10) - (*p2 % 10);
	else
		return *p1 - *p2;
}
int MyCompare3(const void * e1, const void * e2)
{
	A * p1 = (A*)e1;
	A * p2 = (A*)e2;
	return p1->n - p2->n;
}
int a[20];
A b[20];
int main()
{
	freopen("f:\\freopen.txt", "r", stdin);
	int n;
	while (cin >> n) {
		for (int i = 0; i < n; ++i) {
			cin >> a[i];
			b[i].n = a[i];
		}
		mysort(a, n, sizeof(int), MyCompare1);
		for (int i = 0; i < n; ++i)
			cout << a[i] << ",";
		cout << endl;
		mysort(a, n, sizeof(int), MyCompare2);
		for (int i = 0; i < n; ++i)
			cout << a[i] << ",";
		cout << endl;
		mysort(b, n, sizeof(A), MyCompare3);
		for (int i = 0; i < n; ++i)
			cout << b[i].n << ",";
		cout << endl;
	}
	return 0;
}

//*****************This version can pass openjudge test
#include <iostream>
#include <cstring>
using namespace std;

struct Student{
	char name[30];
	int score;
} students[30];
int main()
{
	//freopen("f:\\freopen.txt", "r", stdin);
	int n; 
	cin >> n;
	for(int i = 0;i < n; ++i) 
		cin >> students[i].name >> students[i].score;
	for(int i = n-1; i >= 0; --i) 
		for(int j = 0; j < i; ++ j) 
			if( students[j].score < students[j+1].score ||
				students[j].score == students[j+1].score && 
				strcmp( students[j].name ,students[j+1].name ) > 0) {
					Student tmp;
					tmp = students[j];
					students[j] = students[j+1];
					students[j+1] = tmp;
				}
	for(int i = 0;i < n; ++i)
		cout << students[i].name<< " " << students[i].score<< endl; 
	return 0;
}

//*************************This version can't pass openjudge test but can't figure out
//5.4 2019
#include<iostream>
using namespace std;
struct student
{
	char  name[20 + 10];
	short score;
};
typedef struct student stu;

int compare1(const void * a, const void * b)
{
	return (*(stu *)a).score < (*(stu *)b).score;
}

int compare2(const void * a, const void * b)
{
	if ((*(stu *)a).score == (*(stu *)b).score)
		return (*(stu *)a).name < (*(stu *)b).name;
}

int main()
{
	freopen("f:\\freopen.txt", "r", stdin);
	short num;
	cin >> num;
	stu group[20+5]; 
	for (short i = 0; i < num; i++)
	{
		cin>>group[i].name;
		cin >> group[i].score;
	}
	qsort(group, num, sizeof(stu), compare1);
	qsort(group, num, sizeof(stu), compare2);
	for (short i = 0; i < num; i++)
	{
		cout << group[i].name << " ";
		cout << group[i].score << endl;
	}
	return 0;
}

/*
编写GetDoubleFromString函数，该函数可以不断从字符串中取出非负浮点数，再无浮点数可取，则返回值小于0 
	
输入样例
please 121a1 stand 0.7 9.2 1010.3983 0.00001 black stand what 1324.3
12.34 45 78ab78.34

输出样例
121.000000
1.000000
0.700000
9.200000
1010.398300
0.000010
1324.300000
12.340000
45.000000
78.000000
78.340000
*/	

#include <iostream>
#include <iomanip>
using namespace std;
double GetDoubleFromString(char * str)
{	
	//your code starts here
	static char * p;
	if( str ) 
		p = str;
	double num = 0;

	while( *p && !(*p >= '0' && *p <= '9'))
		++p;
	if( *p == 0)
		return -1;
	while( *p >= '0' && *p <= '9' ) {
		num = num * 10 + *p - '0';
		++p;
	}
	if( *p == '.') {
		++p;
		double i = 10;
		while( *p >= '0' && *p <= '9' ) {
			num += (*p-'0') / i;
			++p;
			i*=10;
		}
	}
	return num;
//your code ends here	
}

int main()
{
	char line[300];
	while(cin.getline(line,280)) {
		double n;
		n = GetDoubleFromString(line);
		while( n > 0) {
			cout << fixed << setprecision(6) << n << endl;
			n = GetDoubleFromString(NULL);
		}
	}
	return 0;
}