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TimeLimited。逐字节输入，维护输出队列，然后对队列快排。讲真，这题测试数据是针对快排的吗？？在下不才，感觉理论上已经是最优解了

Posted by Sunshone at 2018-07-18 20:45:50 on Problem 1002

#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAXS 10000000
int n, leno, flag, book[MAXS] = { 0 }, outlist[1000010] = { 0 }, t[500] = { 0 };
char line[20] = { 0 }, ch;
int out[10];
void Input()
{
	int i, Tel, counter;
	for (i = 0; i <= 9; i++)	t[i+48] = i;
	t['A'] = t['B'] = t['C'] = 2;
	t['D'] = t['E'] = t['F'] = 3;
	t['G'] = t['H'] = t['I'] = 4;
	t['J'] = t['K'] = t['L'] = 5;
	t['M'] = t['N'] = t['O'] = 6;
	t['P'] = t['R'] = t['S'] = 7;
	t['T'] = t['U'] = t['V'] = 8;
	t['W'] = t['X'] = t['Y'] = 9;
	cin >> n;
	flag = leno = 0;
	for (i = 0; i < n; i++)
	{
		counter = 0;
		for (counter = Tel = 0; counter < 7; counter++)
		{
			cin >> ch;
			if (ch == '-' || ch == 'Q' || ch == 'Z'){
				counter--;
				continue;
			}
			Tel = Tel * 10 + t[ch];
		}
		if (Tel > MAXS){		// 异常值保护模块
			printf("ALERT\n"); 
			system("pause");
		}
		flag += (book[Tel] == 0);
		if (book[Tel] == 1) outlist[leno++] = Tel;		//	book[]： Tel 的 frequency，outlist[]堆栈，有值
		book[Tel]++;
	}
}
void FormOut(int a)
{
	int i;
	memset(out, 0, sizeof(out));
	for (i = 1000000; i >= 10000; i /= 10)
		printf("%d", (a % (i * 10)) / i);
	printf("-");
	for(;i > 1; i/=10)
		printf("%d", (a % (i * 10)) / i);
	printf("%d %d\n", a%10, book[a]);
}
void Output()
{
	int i;
	if (flag == n){
		printf("No duplicates.\n");
		return;
	}
	for (i = 0; i < leno; i++)
		FormOut(outlist[i]);
		//printf("%d-%d %d\n", outlist[i]/10000, outlist[i]%10000, book[outlist[i]]);
}
void test()
{
	int i;
	for (i = 0; i < leno; i++)
		printf("%d# MultiplePhoneNum is %d, frequence is %d\n", i, outlist[i], book[outlist[i]]);
	printf("leno = %d\n", leno);
	system("pause");
}
int main()
{
	Input();
	//qsort(outlist, leno, sizeof(outlist[0]), cmp);
	sort(outlist, outlist + leno);
	//test();
	
	Output();
	//system("pause");
	return 0;
}

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Content:

> #include<iostream>
> #include<string.h>
> #include<algorithm>
> using namespace std;
> #define MAXS 10000000
> int n, leno, flag, book[MAXS] = { 0 }, outlist[1000010] = { 0 }, t[500] = { 0 };
> char line[20] = { 0 }, ch;
> int out[10];
> void Input()
> {
> 	int i, Tel, counter;
> 	for (i = 0; i <= 9; i++)	t[i+48] = i;
> 	t['A'] = t['B'] = t['C'] = 2;
> 	t['D'] = t['E'] = t['F'] = 3;
> 	t['G'] = t['H'] = t['I'] = 4;
> 	t['J'] = t['K'] = t['L'] = 5;
> 	t['M'] = t['N'] = t['O'] = 6;
> 	t['P'] = t['R'] = t['S'] = 7;
> 	t['T'] = t['U'] = t['V'] = 8;
> 	t['W'] = t['X'] = t['Y'] = 9;
> 	cin >> n;
> 	flag = leno = 0;
> 	for (i = 0; i < n; i++)
> 	{
> 		counter = 0;
> 		for (counter = Tel = 0; counter < 7; counter++)
> 		{
> 			cin >> ch;
> 			if (ch == '-' || ch == 'Q' || ch == 'Z'){
> 				counter--;
> 				continue;
> 			}
> 			Tel = Tel * 10 + t[ch];
> 		}
> 		if (Tel > MAXS){		// 异常值保护模块
> 			printf("ALERT\n"); 
> 			system("pause");
> 		}
> 		flag += (book[Tel] == 0);
> 		if (book[Tel] == 1) outlist[leno++] = Tel;		//	book[]： Tel 的 frequency，outlist[]堆栈，有值
> 		book[Tel]++;
> 	}
> }
> void FormOut(int a)
> {
> 	int i;
> 	memset(out, 0, sizeof(out));
> 	for (i = 1000000; i >= 10000; i /= 10)
> 		printf("%d", (a % (i * 10)) / i);
> 	printf("-");
> 	for(;i > 1; i/=10)
> 		printf("%d", (a % (i * 10)) / i);
> 	printf("%d %d\n", a%10, book[a]);
> }
> void Output()
> {
> 	int i;
> 	if (flag == n){
> 		printf("No duplicates.\n");
> 		return;
> 	}
> 	for (i = 0; i < leno; i++)
> 		FormOut(outlist[i]);
> 		//printf("%d-%d %d\n", outlist[i]/10000, outlist[i]%10000, book[outlist[i]]);
> }
> void test()
> {
> 	int i;
> 	for (i = 0; i < leno; i++)
> 		printf("%d# MultiplePhoneNum is %d, frequence is %d\n", i, outlist[i], book[outlist[i]]);
> 	printf("leno = %d\n", leno);
> 	system("pause");
> }
> int main()
> {
> 	Input();
> 	//qsort(outlist, leno, sizeof(outlist[0]), cmp);
> 	sort(outlist, outlist + leno);
> 	//test();
> 	
> 	Output();
> 	//system("pause");
> 	return 0;
> }

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