编程题1：
描述
写一个MyString 类，使得下面程序的输出结果是：
1. abcd-efgh-abcd-
2. abcd-
3.
4. abcd-efgh-
5. efgh-
6. c
7. abcd-
8. ijAl-
9. ijAl-mnop
10. qrst-abcd-
11. abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-
要求：MyString类必须是从C++的标准类string类派生而来。提示1：如果将程序中所有 "MyString" 用"string" 替换，那么题目的程序中除了最后两条语句编译无法通过外，其他语句都没有问题，而且输出和前面给的结果吻合。也就是说，MyString类对 string类的功能扩充只体现在最后两条语句上面。提示2: string类有一个成员函数 string substr(int start,int length); 能够求从 start位置开始，长度为length的子串
程序：
#include <iostream>
#include <cstring>
#include <cstdlib>
using namespace std;// 在此处补充你的代码int CompareString( const void * e1, const void * e2) {
MyString * s1 = (MyString * ) e1;
MyString * s2 = (MyString * ) e2;
if( *s1 < *s2 ) return -1;
else if( *s1 == *s2 ) return 0;
else if( *s1 > *s2 ) return 1;
}
int main() {
MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);
MyString SArray[4] = {"big","me","about","take"};
cout << "1. " << s1 << s2 << s3<< s4<< endl;
s4 = s3; s3 = s1 + s3;
cout << "2. " << s1 << endl;
cout << "3. " << s2 << endl;
cout << "4. " << s3 << endl;
cout << "5. " << s4 << endl;
cout << "6. " << s1[2] << endl;
s2 = s1; s1 = "ijkl-";
s1[2] = 'A' ;
cout << "7. " << s2 << endl;
cout << "8. " << s1 << endl;
s1 += "mnop";
cout << "9. " << s1 << endl;
s4 = "qrst-" + s2;
cout << "10. " << s4 << endl;
s1 = s2 + s4 + " uvw " + "xyz";
cout << "11. " << s1 << endl;
qsort(SArray,4,sizeof(MyString), CompareString);
for( int i = 0;i < 4;++i )
cout << SArray[i] << endl;
//输出s1从下标0开始长度为4的子串
cout << s1(0,4) << endl;
//输出s1从下标为5开始长度为10的子串
cout << s1(5,10) << endl;
return 0;
}输入
无
输出
1. abcd-efgh-abcd-
2. abcd-
3.
4. abcd-efgh-
5. efgh-
6. c
7. abcd-
8. ijAl-
9. ijAl-mnop
10. qrst-abcd-
11. abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-
样例输入
无样例输出
1. abcd-efgh-abcd-
2. abcd-
3.
4. abcd-efgh-
5. efgh-
6. c
7. abcd-
8. ijAl-
9. ijAl-mnop
10. qrst-abcd-
11. abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-程序：
#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
class MyString : public string {
public:
MyString() :string() {}; //调用string的构造函数，实际上无参数
MyString(const char * s) :string(s) {}; //调用string构造函数，传入参数s
MyString(const string & s) : string(s) {}; //同上，只不过MyString类的构造函数的形参变为const string &s
MyString operator() (int b, int e) { //()运算符重载，返回MyString类的（s,l)的字符串
return this->substr(b, e);
}
};
int main()
{
MyString s1("abcd-"), s2, s3("efgh-"), s4(s1);
MyString SArray[4] = { "big","me","about","take" };
cout << "1. " << s1 << s2 << s3 << s4 << endl; //输出1.abcd-efgh-abcd-
s4 = s3; //传值，不改变地址
s3 = s1 + s3;
cout << "2. " << s1 << endl; //输出2.abcd-
cout << "3. " << s2 << endl; //输出3.
cout << "4. " << s3 << endl; //输出4.abcd-efgh-
cout << "5. " << s4 << endl; //输出5.efgh-
cout << "6. " << s1[2] << endl; //输出6.c
s2 = s1;
s1 = "ijkl-";
s1[2] = 'A';
cout << "7. " << s2 << endl; //输出7.abcd-
cout << "8. " << s1 << endl; //输出8.ijAl-
s1 += "mnop";
cout << "9. " << s1 << endl; //输出9.ijAl-mnop
s4 = "qrst-" + s2;
cout << "10. " << s4 << endl; //输出10.qrst-abcd-
s1 = s2 + s4 + " uvw " + "xyz";
cout << "11. " << s1 << endl; //输出11.abcd-qrst-abcd- uvw xyz
sort(SArray, SArray + 4); //将SArray排序
for (int i = 0; i < 4; i++)
cout << SArray[i] << endl; //将排序完的数组输出
//s1的从下标0开始长度为4的子串
cout << s1(0, 4) << endl; //()已经经过重载
//s1的从下标5开始长度为10的子串
cout << s1(5, 10) << endl;
system("pause");
return 0;
}编程题2：
描述
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、生命值这两种属性。
有的武士可以拥有武器。武器有三种，sword, bomb,和arrow，编号分别为0,1,2。
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。
不同的武士有不同的特点。
dragon 可以拥有一件武器。编号为n的dragon降生时即获得编号为 n%3 的武器。dragon还有“士气”这个属性，是个浮点数，其值为它降生后其司令部剩余生命元的数量除以造dragon所需的生命元数量。
ninja可以拥有两件武器。编号为n的ninja降生时即获得编号为 n%3 和 (n+1)%3的武器。
iceman有一件武器。编号为n的iceman降生时即获得编号为 n%3 的武器。
lion 有“忠诚度”这个属性，其值等于它降生后其司令部剩余生命元的数目。
wolf没特点。
请注意，在以后的题目里，武士的士气，生命值，忠诚度在其生存期间都可能发生变化，都有作用，武士手中的武器随着使用攻击力也会发生变化。
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。
如果造出的是dragon，那么还要输出一行，例：
It has a arrow,and it's morale is 23.34
表示该dragon降生时得到了arrow,其士气是23.34（为简单起见，本题中arrow前面的冠词用a,不用an，士气精确到小数点后面2位，四舍五入）
如果造出的是ninja，那么还要输出一行，例：
It has a bomb and a arrow
表示该ninja降生时得到了bomb和arrow。
如果造出的是iceman，那么还要输出一行，例：
It has a sword
表示该iceman降生时得到了sword。
如果造出的是lion，那么还要输出一行，例：
It's loyalty is 24
表示该lion降生时的忠诚度是24。
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。
输入
第一行是一个整数,代表测试数据组数。
每组测试数据共两行。
第一行，一个整数M。其含义为： 每个司令部一开始都有M个生命元( 1 <= M <= 10000)
第二行：五个整数，依次是 dragon 、ninja、iceman、lion、wolf 的初始生命值。它们都大于0小于等于10000
输出
对每组测试数据，要求输出从0时0分开始，到双方司令部都停止制造武士为止的所有事件。
对每组测试数据，首先输出“Case:n" n是测试数据的编号，从1开始
接下来按恰当的顺序和格式输出所有事件。每个事件都以事件发生的时间开头，时间以小时为单位，有三位。
样例输入
1
20
3 4 5 6 7样例输出
Case:1
000 red iceman 1 born with strength 5,1 iceman in red headquarter
It has a bomb
000 blue lion 1 born with strength 6,1 lion in blue headquarter
It's loyalty is 14
001 red lion 2 born with strength 6,1 lion in red headquarter
It's loyalty is 9
001 blue dragon 2 born with strength 3,1 dragon in blue headquarter
It has a arrow,and it's morale is 3.67
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
It has a sword and a bomb
003 red headquarter stops making warriors
003 blue iceman 4 born with strength 5,1 iceman in blue headquarter
It has a bomb
004 blue headquarter stops making warriors程序：
#include <iostream>
#include <cstdio>
#include <string>
#include <iomanip>
using namespace std;
const int WARRIOR_NUM = 5;
class Headquarter;
class Warrior
{
private:
Headquarter * pHeadquarter;
int kindNo;
int no;
public:
static string names[WARRIOR_NUM];
static int initialLifeValue[WARRIOR_NUM];
static string weaponName[3];
Warrior(Headquarter * p, int no_, int kindNo_);
void PrintResult(int nTime);
};
class Headquarter
{
private:
int totalLifeValue;
bool stopped;
int totalWarriorNum;
int color;
int warriorNum[WARRIOR_NUM];
Warrior * pWarriors[1000];
public:
friend class Warrior;
int curMakingSeqIdx;
static int makingSeq[2][WARRIOR_NUM];
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());
if (pHeadquarter->color == 0 && pHeadquarter->curMakingSeqIdx == 1) cout << "It has a " << weaponName[pHeadquarter->curMakingSeqIdx % 3] << endl;
if (pHeadquarter->color == 1 && pHeadquarter->curMakingSeqIdx == 1) cout << "It's loyalty is " << pHeadquarter->totalLifeValue << endl;
if (pHeadquarter->color == 0 && pHeadquarter->curMakingSeqIdx == 2) cout << "It's loyalty is " << pHeadquarter->totalLifeValue << endl;
if (pHeadquarter->color == 1 && pHeadquarter->curMakingSeqIdx == 2) { cout << "It has a " << weaponName[pHeadquarter->curMakingSeqIdx % 3] << ",and it's morale is "; cout << fixed << setprecision(2) << float(pHeadquarter->totalLifeValue) / float(initialLifeValue[0]) << endl; }
//if (pHeadquarter->color == 0 && pHeadquarter->curMakingSeqIdx == 1) cout << "It has a " << weaponName[pHeadquarter->totalWarriorNum % 3] << "\n" << endl;
if (pHeadquarter->color == 1 && pHeadquarter->curMakingSeqIdx == 3) cout << "It has a " << weaponName[pHeadquarter->curMakingSeqIdx % 3] << " and a " << weaponName[(pHeadquarter->curMakingSeqIdx + 1) % 3] << endl;
if (pHeadquarter->color == 0 && pHeadquarter->curMakingSeqIdx == 4) cout << "It has a " << weaponName[pHeadquarter->curMakingSeqIdx % 3] << " and a " << weaponName[(pHeadquarter->curMakingSeqIdx + 1) % 3] << endl;
if (pHeadquarter->color == 1 && pHeadquarter->curMakingSeqIdx == 4) cout << "It has a " << weaponName[pHeadquarter->curMakingSeqIdx % 3] << endl;
if (pHeadquarter->color == 0 && pHeadquarter->curMakingSeqIdx == 5) cout << "It has a " << weaponName[pHeadquarter->curMakingSeqIdx % 3] << ",and it's morale is" << float(pHeadquarter->totalLifeValue) / float(initialLifeValue[1]) << endl;
////if (pHeadquarter->color == 1 && pHeadquarter->curMakingSeqIdx == 1) cout << "It's loyalty is" << pHeadquarter->totalLifeValue << "\n" << endl;
}
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" };
string Warrior::weaponName[3] = { "sword", "bomb", "arrow" };
int Warrior::initialLifeValue[WARRIOR_NUM];
int Headquarter::makingSeq[2][WARRIOR_NUM] = { { 2, 3, 4, 1, 0 },{ 3, 0, 1, 2, 4 } };
int main()
{
int t;
int m;
Headquarter RedHead, BlueHead;
cin >> t;
int nCaseNo = 1;
while (t--) {
printf("Case:%d\n", nCaseNo++);
scanf_s("%d", &m,3);
for (int i = 0; i < WARRIOR_NUM; i++)
scanf_s("%d", &Warrior::initialLifeValue[i],6);
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++;
}
}
system("pause");
return 0;
}