学校的C++讲的实在太扯,于是只能在网上自学。这是网上的课后编程作业,看后着实吓了一跳,学编程半年多了还没见过这么NB的程序题。。。
先粘题目:
魔兽世界的西面是红魔军的司令部,东面是蓝魔军的司令部。两个司令部之间是依次排列的若干城市。
红司令部,City 1,City 2,……,City n,蓝司令部
两军的司令部都会制造武士。武士一共有 dragon 、ninja、iceman、lion、wolf 五种。每种武士都有编号、生命值、攻击力这三种属性。
双方的武士编号都是从1开始计算。红方制造出来的第n个武士,编号就是n。同样,蓝方制造出来的第n个武士,编号也是n。
武士在刚降生的时候有一个生命值。
在每个整点,双方的司令部中各有一个武士降生。
红方司令部按照iceman、lion、wolf、ninja、dragon的顺序循环制造武士。
蓝方司令部按照lion、dragon、ninja、iceman、wolf的顺序循环制造武士。
制造武士需要生命元。
制造一个初始生命值为m的武士,司令部中的生命元就要减少m个。
如果司令部中的生命元不足以制造某个按顺序应该制造的武士,那么司令部就试图制造下一个。如果所有武士都不能制造了,则司令部停止制造武士。
给定一个时间,和双方司令部的初始生命元数目,要求你将从0点0分开始到双方司令部停止制造武士为止的所有事件按顺序输出。
一共有两种事件,其对应的输出样例如下:
1) 武士降生
输出样例: 004 blue lion 5 born with strength 5,2 lion in red headquarter
表示在4点整,编号为5的蓝魔lion武士降生,它降生时生命值为5,降生后蓝魔司令部里共有2个lion武士。(为简单起见,不考虑单词的复数形式)注意,每制造出一个新的武士,都要输出此时司令部里共有多少个该种武士。
2) 司令部停止制造武士
输出样例: 010 red headquarter stops making warriors
表示在10点整,红方司令部停止制造武士
输出事件时:
首先按时间顺序输出;
同一时间发生的事件,先输出红司令部的,再输出蓝司令部的。
1 20 3 4 5 6 7
Case:1 000 red iceman 1 born with strength 5,1 iceman in red headquarter 000 blue lion 1 born with strength 6,1 lion in blue headquarter 001 red lion 2 born with strength 6,1 lion in red headquarter 001 blue dragon 2 born with strength 3,1 dragon in blue headquarter 002 red wolf 3 born with strength 7,1 wolf in red headquarter 002 blue ninja 3 born with strength 4,1 ninja in blue headquarter 003 red headquarter stops making warriors 003 blue iceman 4 born with strength 5,1 iceman in blue headquarter 004 blue headquarter stops making warriors
只能说做这题的时候面向对象的思想还是一团浆糊,就这么比葫芦画瓢的写起来。
类没有设计好,写代码的过程真是非常困难啊。(以后一定要把类设计好再写~~~~(>_<)~~~~ )
另外也遇到了不少问题,比如“如果司令部中的生命元不足以制造某个按顺序应该制造的武士,那么司令部就试图制造下一个”这个问题,就纠结了好长时间,最后交了个很脑残的处理方法。现在想想还是当时太仓促没有全盘考虑。。。
然后经过无比**的修修补补,总算及时交了程序,但是写的自己都看不下去(print竟然还是在main里实现的→_→):
#include
#include
#include
#include
#include
using namespace std;
#define MAX 99999999
class Headquarter;
class Sumurai; //声明类
int M = 0; //M初始生命元
int gametime; //游戏时间
int key0,key1; //制造结束标志,0为红,1为蓝
int Num[2]; //各司令部武士数量,0为红,1为蓝
int Next[2]; //记录下一个要建造的武士在order中的位置,0为红,1为蓝
int order[2][5]={{2,3,4,1,0},{3,0,1,2,4}}; //dragon、ninja、iceman、lion、wolf分别用0~4代替
int original_elem[5]={0,0,0,0,0}; //记录每种武士的生命元
char name[5][7] = {"dragon","ninja","iceman","lion","wolf"}; //武士名称
char headname[2][5]={"red","blue"}; //阵营名称
Headquarter* head[2]; //指向两司令部,0为红,1为蓝
Sumurai* sumu[2][1000]; //武士的记录,0为红,1为蓝
class Headquarter{
public:
int elem; //剩余生命元
int color; //属于哪个阵营,红为0,蓝为1
int count[5]; //每种武士的数量
Headquarter(int a){
elem = M;
color = a;
memset(count,0,sizeof(count));
}//构造函数初始化
bool GiveBirth() //制造武士
{
int sort = order[color][Next[color]%5]; //当前应制造的武士类型
if(elem - original_elem[sort] >= 0 )
{
elem -= original_elem[sort]; //扣除生命元制造武士
count[sort]++;
printf("%03d %s %s %d born with strength %d,%d %s in %s headquarter\n",
gametime/60,headname[color],name[sort],Num[color],original_elem[sort], count[sort],name[sort],headname[color]);
Num[color]++;
Next[color]++;
return true;
}
else if(elem >= original_elem[order[color][(Next[color]+1) % 5]]){
sort = order[color][(Next[color]+1)%5];
elem -= original_elem[sort]; //扣除生命元制造武士
count[sort]++;
printf("%03d %s %s %d born with strength %d,%d %s in %s headquarter\n",
gametime/60,headname[color],name[sort],Num[color],original_elem[sort], count[sort],name[sort],headname[color]);
Num[color]++;
Next[color] += 2;
return true;
}
else if(elem >= original_elem[order[color][(Next[color]+2)%5]]){
sort = order[color][(Next[color]+2)%5];
elem -= original_elem[sort]; //扣除生命元制造武士
count[sort]++;
printf("%03d %s %s %d born with strength %d,%d %s in %s headquarter\n",
gametime/60,headname[color],name[sort],Num[color],original_elem[sort], count[sort],name[sort],headname[color]);
Num[color]++;
Next[color] += 3;
return true;
}
else if(elem >= original_elem[order[color][(Next[color]+3)%5]]){
sort = order[color][(Next[color]+3)%5];
elem -= original_elem[sort]; //扣除生命元制造武士
count[sort]++;
printf("%03d %s %s %d born with strength %d,%d %s in %s headquarter\n",
gametime/60,headname[color],name[sort],Num[color],original_elem[sort], count[sort],name[sort],headname[color]);
Num[color]++;
Next[color] += 4;
return true;
}
else if(elem >= original_elem[order[color][(Next[color]+4)%5]]){
sort = order[color][(Next[color]+4)%5];
elem -= original_elem[sort]; //扣除生命元制造武士
count[sort]++;
printf("%03d %s %s %d born with strength %d,%d %s in %s headquarter\n",
gametime/60,headname[color],name[sort],Num[color],original_elem[sort], count[sort],name[sort],headname[color]);
Num[color]++;
Next[color] += 5;
return true;
}
else return false;
}
};//司令部类
class Sumurai
{
};//武士类
int main()
{
//freopen("in.txt","r",stdin);
//freopen("myout.txt","w",stdout);
int test,i_test;
cin>>test;
for(i_test=1; i_test<=test; i_test++)
{
Num[0] = Num[1] = 1; //各司令部武士数量置零
key0 = key1 = 0; //结束标志置零
Next[0] = Next[1] = 0; //要建造的武士在order中的位置置零
cin>>M;//读入初始生命元
for(int i=0;i<5;i++)
cin>>original_elem[i]; //读入每种武士的生命元
head[0] = new Headquarter(0); //产生红司令部
head[1] = new Headquarter(1); //产生蓝司令部
printf("Case:%d\n",i_test);
for(gametime = 0; key0==0 || key1==0; gametime++)
{
if(gametime % 60 == 0)
{
if(head[0]->GiveBirth() == false){
if(key0 == 0)
printf("%03d red headquarter stops making warriors\n",gametime/60);
key0 = 1;
}
if(head[1]->GiveBirth() == false){
if(key1 == 0)
printf("%03d blue headquarter stops making warriors\n",gametime/60);
key1 = 1;
}
}
}
}
//system("pause");
return 0;
}
然后这是我觉得写的很好的老师的代码(至于哪里好就不列举了,因为哪里都好):
#include
#include
#include
using namespace std;
const int WARRIOR_NUM = 5;
/*
string Warrior::names[WARRIOR_NUM] = { "dragon","ninja","iceman","lion","wolf" };
红方司令部按照 iceman、lion、wolf、ninja、dragon 的顺序制造武士。
蓝方司令部按照 lion、dragon、ninja、iceman、wolf 的顺序制造武士。
*/
class Headquarter;
class Warrior
{
private:
Headquarter * pHeadquarter;
int kindNo; //武士的种类编号 0 dragon 1 ninja 2 iceman 3 lion 4 wolf
int no;
public:
static string names[WARRIOR_NUM];
static int initialLifeValue [WARRIOR_NUM];
Warrior( Headquarter * p,int no_,int kindNo_ );
void PrintResult(int nTime);
};
class Headquarter
{
private:
int totalLifeValue;
bool stopped;
int totalWarriorNum;
int color;
int curMakingSeqIdx; //当前要制造的武士是制造序列中的第几个
int warriorNum[WARRIOR_NUM]; //存放每种武士的数量
Warrior * pWarriors[1000];
public:
friend class Warrior;
static int makingSeq[2][WARRIOR_NUM]; //武士的制作顺序序列
void Init(int color_, int lv);
~Headquarter () ;
int Produce(int nTime);
string GetColor();
};
Warrior::Warrior( Headquarter * p,int no_,int kindNo_ ) {
no = no_;
kindNo = kindNo_;
pHeadquarter = p;
}
void Warrior::PrintResult(int nTime)
{
string color = pHeadquarter->GetColor();
printf("%03d %s %s %d born with strength %d,%d %s in %s headquarter\n" ,
nTime, color.c_str(), names[kindNo].c_str(), no, initialLifeValue[kindNo],
pHeadquarter->warriorNum[kindNo],names[kindNo].c_str(),color.c_str());
}
void Headquarter::Init(int color_, int lv)
{
color = color_;
totalLifeValue = lv;
totalWarriorNum = 0;
stopped = false;
curMakingSeqIdx = 0;
for( int i = 0;i < WARRIOR_NUM;i++ )
warriorNum[i] = 0;
}
Headquarter::~Headquarter () {
for( int i = 0;i < totalWarriorNum;i++ )
delete pWarriors[i];
}
int Headquarter::Produce(int nTime)
{
if( stopped )
return 0;
int searchingTimes = 0;
while( Warrior::initialLifeValue[makingSeq[color][curMakingSeqIdx]] > totalLifeValue &&
searchingTimes < WARRIOR_NUM ) {
curMakingSeqIdx = ( curMakingSeqIdx + 1 ) % WARRIOR_NUM;
searchingTimes++;
}
int kindNo = makingSeq[color][curMakingSeqIdx];
if( Warrior::initialLifeValue[kindNo] > totalLifeValue ) {
stopped = true;
if( color == 0)
printf("%03d red headquarter stops making warriors\n",nTime);
else
printf("%03d blue headquarter stops making warriors\n",nTime);
return 0;
}
//制作士兵:
totalLifeValue -= Warrior::initialLifeValue[kindNo];
curMakingSeqIdx = ( curMakingSeqIdx + 1 ) % WARRIOR_NUM;
pWarriors[totalWarriorNum] = new Warrior( this,totalWarriorNum+1,kindNo);
warriorNum[kindNo]++;
pWarriors[totalWarriorNum]->PrintResult(nTime);
totalWarriorNum++;
return 1;
}
string Headquarter::GetColor()
{
if( color == 0)
return "red";
else
return "blue";
}
string Warrior::names[WARRIOR_NUM] = { "dragon","ninja","iceman","lion","wolf" };
int Warrior::initialLifeValue [WARRIOR_NUM];
int Headquarter::makingSeq[2][WARRIOR_NUM] = { { 2,3,4,1,0 },{3,0,1,2,4} }; //两个司令部武士的制作顺序序列
int main()
{
int t;
int m;
Headquarter RedHead,BlueHead;
scanf("%d",&t);
int nCaseNo = 1;
while ( t-- ) {
printf("Case:%d\n",nCaseNo++);
scanf("%d",&m);
for( int i = 0;i < WARRIOR_NUM;i++ )
scanf("%d", & Warrior::initialLifeValue[i]);
RedHead.Init(0,m);
BlueHead.Init(1,m);
int nTime = 0;
while(true) {
int tmp1 = RedHead.Produce(nTime);
int tmp2 = BlueHead.Produce(nTime);
if( tmp1 == 0 && tmp2 == 0)
break;
nTime++;
}
}
return 0;
}