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本文总结了TopCoder SRM比赛中的多个编程题目解决思路，涵盖了时间转换、二进制解码、最大销售额计算等典型算法问题。通过具体代码示例介绍了如何使用C++进行字符串操作、类型转换及高效算法设计。

SRM 144 DIV2

1.*200 --- 60 read

#include<string>
#include<iostream>
#include<sstream>
using namespace std;
class Time {
public:
	string whatTime(int seconds);
private:
	int hr,min,sec;
};

string Time::whatTime (int seconds) {
	hr=seconds/3600;
	min=(seconds-hr*3600)/60;
	sec=seconds-hr*3600-min*60;
	stringstream ss;
	ss << hr << ":" << min << ":" << sec;
	return ss.str();
}

How to convert int into string. Important!

2.*550 --- 168 read

#include<string>
#include<vector>
#include<iostream>
using namespace std;
class BinaryCode {
public:
	vector<string> decode(string message) {
		int n=message.size();
		string sa="0", sb="1";
		for(int i=0; i<n-1; ++i) {
			if(i==0) sa += message[0];
			else     sa += message[i] - sa[i] - (sa[i-1]-'0')+'0';
			if(sa[i+1]<'0'||sa[i+1]>'1'){
				sa="NONE";
				break;
			}
		}
		for(int i=0; i<n-1; ++i) {
			if(i==0) sb += message[0]-1;
			else     sb += message[i] - sb[i] - (sb[i-1]-'0')+'0';
			if(sb[i+1]<'0'||sb[i+1]>'1'){
				sb="NONE";
				break;
			}
		}
		if(message[n-1]+'0'-sa[n-1]-sa[n-2]!=0) sa="NONE";
		if(message[n-1]+'0'-sb[n-1]-sb[n-2]!=0) sb="NONE";
		vector<string> result;
		result.push_back(sa);
		result.push_back(sb);
		return result;
	}
};

The return value of [] is a char, remember how to calc char with int.

SRM 145 DIV2

1.*250 --- 75

SRM 146 DIV2

1.*250 --- 118

2.*500 --- 461

SRM 147 DIV2

1.*250 --- 241

SRM 147 DIV1

1.*500 --- 195 read

#include <iostream>
#include <vector>
#include <sstream>
#include <string>
using namespace std;

 class Dragons
 {
 public:
 	string snaug(vector <int> initialFood, int rounds)
 	{
 		long long denominator = 4;
 		long long Fn[6];
 		long long Fo[6]={initialFood[0],initialFood[1],initialFood[2],initialFood[3],initialFood[4],initialFood[5]};
 		Fn[0]=Fn[1]=Fo[2]+Fo[3]+Fo[4]+Fo[5];
 		Fn[2]=Fn[3]=Fo[0]+Fo[1]+Fo[4]+Fo[5];
 		Fn[4]=Fn[5]=Fo[0]+Fo[1]+Fo[2]+Fo[3];
 		--rounds;
 		for(int i=1;i<=rounds;++i)
 		{
 			denominator *=2;
 			for(int j=0;j<5;j+=2)
 				Fo[j]=Fn[j];
 			Fn[0]=Fo[2]+Fo[4];
 			Fn[2]=Fo[0]+Fo[4];
 			Fn[4]=Fo[0]+Fo[2];
 			
//Try to divide by 2 to save time
                        if(Fn[0]%2==0 && Fn[2]%2==0 && Fn[4]%2==0)
 			{
 				Fn[0]/=2;
 				Fn[2]/=2;
 				Fn[4]/=2;
 				denominator/=2;
 			}
 		}

//calc final result
 		long long nominator=Fn[2];
 		long long a,b,tmp;
 		a=nominator;
 		b=denominator;
		while(a%b!=0)
		{
			tmp=a%b;
			a=b;
			b=tmp;
		}
		nominator /= b;
		denominator /= b;
// convert from int to string
                stringstream ss;
 		if(nominator%denominator==0)
 			ss<<(nominator/denominator);
 		else
 			ss << nominator << "/" << denominator;
 		return ss.str();
 	}
 };

1. should use "long long" when the value may exceed the boundary

2. should try to divide by 2 to save time!

3. convert from int to string

SRM 148 DIV2

1.*250 --- 81

SRM 149 DIV2

1.*500 --- 451

2.*1000 --- 405 read

#include <iostream>
#include <vector>
using namespace std;

class Pricing
{
public:
	int maxSales(vector <int> price)
	{
		int Max = 0;
		// corner case
		if (price.size()<=4)
		{
			for(vector<int>::iterator iter = price.begin(); iter!=price.end();++iter)
				Max += *iter;
			return Max;
		}
		// valid case
		vector<int> sprice;
		int TotalNum = price.size();
		for(int i=0; i<TotalNum;++i)
		{
		//Min should be 1000, not 0!!!
			int Min=1000;
			vector<int>::iterator Miter;
			for(vector<int>::iterator iter = price.begin(); iter!=price.end();++iter)
			{
				if(Min>*iter)
				{
					Min = *iter;
					Miter = iter;
				}
			}
			sprice.push_back(Min);
			price.erase(Miter);
		}
		int tmp;
		for(int i=0;i<=TotalNum;++i)
			for(int j=i;j<TotalNum; ++j)
				for(int k=j;k<TotalNum; ++k)
					for(int l=k;l<TotalNum; ++l)
					{
						tmp = sprice[i]*(j-i) + sprice[j]*(k-j)+sprice[k]*(l-k)+sprice[l]*(TotalNum-l);
						if(tmp > Max)
							Max = tmp;
					}
		return Max;
	}
};

When try to find the Max, we should set the initial value of Max to minimum value;

When try to find the Min, we should set the initial value of Min to maximum value;

SRM 150 DIV2

1.*250 --- 219

2.*500 --- 435

SRM 151 DIV2

1.*250 --- 187

2.*500 --- 384

SRM 152 DIV2

1.*250 --- 234

2.*500 --- 409

3.*1000 --- 490

SRM 153 DIV2

1.*250 --- 228

2.*500 --- 370

#include <iostream>
#include <vector>
using namespace std;

class Inventory
{
public:
	int monthlyOrder(vector <int> sales, vector <int> daysAvailable)
	{
		double total=0;
		int size = sales.size();
		int emp = 0;
		for(int i=0;i<size;++i)
		{
			if(daysAvailable[i]!=0)
				total+= sales[i]*30/double(daysAvailable[i]);  // must transfer to double to get double result
			else
				++emp;
		}
		size -= emp;
		if(size !=0 )
		{
			total = total/size;
			int round = total;
			if (total != round)
				++round;
			return round;
		}
		else
			return 0;
	}
};

when trying to increase the accuracy, we must ensure not only the L-value is double, the right side expression should also be double.(when there is int denominator, we should change it into double type to ensure the accuracy)

3.*1000 --- 490

SRM 154 DIV2

1.*300 --- 171.58

#include <iostream>
#include <sstream>
#include <vector>

using namespace std;
class 
ProfitCalculator
{
public:
	int percent(vector <string> items)
	{
		stringstream s;
		double cost, price, Tcost, Tprice;
		Tcost = 0;
		Tprice = 0; 
		for(vector<string>::iterator iter = items.begin(); iter != items.end(); ++iter)
		{
			s << *iter;
			s >> price >> cost;
			Tcost += cost;
			Tprice += price;
			s.str("");    //clear buffer
			s.clear();    //reset error state
		}
		double Result =( Tprice - Tcost )/Tprice;
		return int(Result*100);
	}
};

How to transfer from string to int/double.

When we want to reuse a stringstream object, we need to:

1. clear buffer

2. reset the error state