C语言题目14
//实现输入任意两个若干位的整数，进行加法，计算并返回加和后的结果。
//（注意两个整数可以足够大，以至于double都会溢出。只要系统内存够用，
//此加法都要正常运行）
#include"stdio.h"
#include"stdlib.h"
typedef struct node{
int x;
struct node *next;
struct node *before;
}llnode;
int charToNumber(char c)
{
return c-48;
}
void getNumber(llnode* tail)
{
char c;
llnode *p;
llnode *q;
q = tail;
printf("请输入一个整数， 输入完成后敲击回车:");
while(1)
{
c = getchar();
if(c == '\n')
{
break;
}
else
{
p = (llnode *)malloc(sizeof(llnode));
p->x = charToNumber(c);
p->before = NULL;
if(tail->x == 0)
{
tail->before = p;
p->next = tail;
tail->x = tail->x+1;
}
else
{
q = tail->before;
q->next = p;
p->before = q;
p->next = tail;
tail->before = p;
q = p;
tail->x = tail->x+1;
}
}
}
}
void addNumber(llnode* ctail, llnode* atail, llnode* btail)
{
int i=0;
int alength = atail->x;
int blength = btail->x;
llnode *pshort;
llnode *plong;
llnode *ps;
llnode *pl;
llnode *p;
llnode *q;
int a;
int b;
int nextc;
int thisc;
int d;
if(alength>blength)
{
pshort = btail;
plong = atail;
}
else
{
pshort = atail;
plong = btail;
}
ps = pshort->before;
pl = plong->before;
while(i<pshort->x)
{
thisc = nextc;
if(ps->x + pl->x >=10)
{
nextc = 1;
d = (ps->x + pl->x)%10;
}
else
{
nextc = 0;
d = (ps->x + pl->x);
}
ps = ps->before;
pl = pl->before;
p = (llnode *)malloc(sizeof(llnode));
p->x = d+thisc;
p->before = NULL;
if(ctail->x == 0)
{
ctail->before = p;
p->next = ctail;
ctail->x = ctail->x+1;
q = p;
}
else
{
p->next = q;
q->before = p;
q = p;
ctail->x = ctail->x+1;
}
i = i+1;
}
while(i<plong->x)
{
if(i == pshort->x)
{
p = (llnode *)malloc(sizeof(llnode));
p->x = (pl->x)+nextc;
p->before = NULL;
}
else
{
p = (llnode *)malloc(sizeof(llnode));
p->x = pl->x;
p->before = NULL;
}
p->next = q;
q->before = p;
q = p;
ctail->x = ctail->x+1;
pl = pl->before;
i = i+1;
}
}
void showResult(llnode* ctail)
{
llnode *p;
p = ctail;
while(p->before != NULL)
{
p = p->before;
}
while(p != ctail)
{
printf("%d",p->x);
p = p->next;
}
printf("\n");
}
int main()
{
llnode *atail = (llnode *)malloc(sizeof(llnode));
llnode *btail = (llnode *)malloc(sizeof(llnode));
llnode *ctail = (llnode *)malloc(sizeof(llnode));
llnode *p=atail;
atail->x = 0;
btail->x = 0;
ctail->x = 0;
atail->next = NULL;
btail->next = NULL;
ctail->next = NULL;
getNumber(atail);
getNumber(btail);
addNumber(ctail, atail, btail);
showResult(ctail);
return 1;
}
C++语言题目4
//超市每天卖出的货物假如有egg，apple，pear三种，销售账目存储在了字典里，如下：
//account = [{"egg":[5.00, 3, "s"],
//"apple":[2.50, 5, "s"],
//"pear":[4.50, 2, "s"]},
//{"egg":[6.00, 3, "s"],
//"apple":[3.00, 15, "q"],
//"pear":[6.50, 2, "s"]},
//{"egg":[5.50, 5, "s"],
//"apple":[3.50, 5, "s"],
//"pear":[5.00, 6, "q"]},
//{"egg":[5.00, 3, "s"],
//"apple":[2.50, 6, "s"],
//"pear":[5.50, 2, "q"]},
//{"egg":[5.50, 8, "q"],
//"apple":[3.50, 5, "s"],
//"pear":[5.50, 3, "s"]}]
//每天的账目是列表中的一个元素，包括一个字典，字典的键是货物名称，字典的值的第一个元素是单价，
//第二个元素是数目，第三个元素，如果是s表示该账目没有任何疑问，如果是q表示该账目可能有错误。
//我们要做的是事情是统计每种货物，没有任何疑问的账目的销售总额。也就是对标志位为"s"的账目，
//按照货物名称，计算销售总额。（自己可根据使用的编程语言，确定存储结构，最后结果在屏幕打印
//egg、apple和pear分别的没有任何疑问的帐目的销售总额就可以。）
#include"iostream"
#include"string"
using namespace std;
class Account
{
private:
string name;
double price;
float number;
char flag;
public:
Account(string goodsname, double goodsprice, float goodsnumber, char goodsflag)
{
name = goodsname;
price = goodsprice;
number = goodsnumber;
flag = goodsflag;
}
double getTotalPrice()
{
return price*number;
}
string getName()
{
return name;
}
char getFlag()
{
return flag;
}
};
int main()
{
Account accounts[15] = {
Account("egg",5.00,3,'s'),
Account("apple",2.50,5,'s'),
Account("pear",4.50,2,'s'),
Account("egg",6.00,3,'s'),
Account("apple",3.00,15,'q'),
Account("pear",6.50,2,'s'),
Account("egg",5.50,5,'s'),
Account("apple",3.50,5,'s'),
Account("pear",5.00,6,'q'),
Account("egg",5.00,3,'s'),
Account("apple",2.50,6,'s'),
Account("pear",5.50,2,'q'),
Account("egg",5.50,8,'q'),
Account("apple",3.50,5,'s'),
Account("pear",5.50,3,'s')};
string names[3] = {"egg", "apple", "pear"};
int i=0;
int j;
double count=0;
for(i=0; i<3; i++)
{
count = 0;
for(j=0; j<15; j++)
{
if((accounts[j].getName() == names[i]) && (accounts[j].getFlag() == 's'))
{
count = count + accounts[j].getTotalPrice();
}
}
cout<<names[i]<<"的总价是:"<<count<<endl;
}
return 1;
}
C++语言题目3
//采用面向对象的方法实现一个stack类，操作包括进栈，出栈，取栈顶元素，显示整个栈；
//（只要内存够用，可以无限次进栈）
#include"iostream"
using namespace std;
class Node{
private:
int x;
Node* next;
public:
Node(int data)
{
x = data;
}
void setNext(Node* p)
{
next = p;
}
Node* getNext()
{
return next;
}
int setData(int number)
{
x = number;
}
int getData()
{
return x;
}
~Node(){}
};
class Stack
{
private:
Node *nodelast;
public:
Stack()
{
nodelast = new Node(0);
nodelast->setNext(NULL);
}
void pushData(int x)
{
Node *p = new Node(x);
p->setNext(nodelast->getNext());
nodelast->setNext(p);
nodelast->setData(nodelast->getData()+1);
}
int popData()
{
int value;
int error = 0;
Node *p = nodelast->getNext();
Node *q = p;
if((p != NULL))
{
value = q->getData();
p = p->getNext();
nodelast->setNext(p);
nodelast->setData(nodelast->getData()-1);
delete q;
return value;
}
else
{
throw error;
}
}
int getTopData()
{
int value;
int error = 0;
Node *p = nodelast->getNext();
if((p != NULL))
{
value = p->getData();
return value;
}
else
{
throw error;
}
}
void showStack()
{
int i;
Node *p = nodelast;
cout<<"Top ";
for(i=0; i< nodelast->getData(); i++)
{
p = p->getNext();
cout<< p->getData()<<" ";
}
cout<<"End"<<endl;
}
};
int main()
{
int value;
Stack s;
try
{
cout<<"出栈:"<<endl;
s.popData();
}
catch(int)
{
cout<<"栈为空！"<<endl;
}
cout<<"入栈1,3,2:"<<endl;
s.pushData(1);
s.pushData(3);
s.pushData(2);
s.showStack();
try
{
value = s.popData();
cout<<"出栈:"<<value<<endl;
s.showStack();
value = s.getTopData();
cout<<"当前栈顶元素:"<<value<<endl;
s.showStack();
}
catch(int)
{
cout<<"栈为空！"<<endl;
}
cout<<"入栈3,4,5:"<<endl;
s.pushData(3);
s.pushData(4);
s.pushData(5);
s.showStack();
return 1;
}
C语言题目13
//有两列数字， A{1,2,3,4,5,6,7,8,9} B{10,11,12,13,14,15}
//合并成一个新的数列C，要求所有的奇数排在所有的偶数前面，
//并且奇数要升序，偶数要降序，输出数列C到屏幕上
#include"stdio.h"
int getminodd(int c[], int start, int n)
{
int k;
int minodd = start;
while(c[minodd]%2 == 0)
{
minodd = minodd+1;
if(minodd == n)
{
break;
}
}
if(minodd != n)
{
for(k=minodd; k<n; k++)
{
if((c[minodd]>c[k]) && c[k]%2 == 1)
{
minodd = k;
}
}
}
return minodd;
}
int getmaxeven(int c[], int start, int n)
{
int k;
int maxeven = start;
while(c[maxeven]%2 != 0)
{
maxeven = maxeven+1;
if(maxeven == n)
{
break;
}
}
if(maxeven != n)
{
for(k=maxeven; k<n; k++)
{
if((c[maxeven]<c[k]) && c[k]%2 == 0)
{
maxeven = k;
}
}
}
return maxeven;
}
int main()
{
int a[9] = {1,2,3,4,5,6,7,8,9};
int b[6] = {10,11,12,13,14,15};
int c[15] = {};
int i;
int j;
int k=0;
int count=0;
int minodd=0;
int maxeven=0;
int swap;
//合并
for(i=0; i<9; i++)
{
c[k] = a[i];
k = k+1;
}
for(j=0; j<6; j++)
{
c[k] = b[j];
k = k+1;
}
//奇数部分选择排序
while(1)
{
minodd = getminodd(c, count, 15);
if(minodd == 15)
{
break;
}
swap = c[count];
c[count] = c[minodd];
c[minodd] = swap;
count = count+1;
if(count >= 15)
{
break;
}
}
//偶数部分选择排序
while(1)
{
maxeven = getmaxeven(c, count, 15);
if(maxeven == 15)
{
break;
}
swap = c[count];
c[count] = c[maxeven];
c[maxeven] = swap;
count = count+1;
if(count >= 15)
{
break;
}
}
//输出
for(k=0; k<15; k++)
{
printf("%d ", c[k]);
}
printf("\n");
return 1;
}
C语言题目12
//有两个集合， A{1,3,6,2,9} B{1,4,6,9,10} 输出这两个集合的交集和并集到屏幕上；
#include"stdio.h"
int main()
{
int a[5] = {1,3,6,2,9};
int b[5] = {1,4,6,9,10};
int i;
int j;
//输出交集
printf("两个集合的交集:");
for(i=0; i<5; i++)
{
for(j=0; j<5; j++)
{
if(a[i] == b[j])
{
printf("%d ", a[i]);
break;
}
}
}
printf("\n");
//输出并集
printf("两个集合的并集:");
for(i=0; i<5; i++)
{
for(j=0; j<5; j++)
{
if(a[i] == b[j])
{
a[i] = '#';
}
}
}
for(i=0; i<5; i++)
{
if(a[i]!='#')
{
printf("%d ", a[i]);
}
}
for(j=0; j<5; j++)
{
printf("%d ", b[j]);
}
printf("\n");
return 1;
}
C语言题目11
//输入一串字符串，以回车结束，统计其中的英文字符、数字字符、空格和其他字符数目；
#include<stdio.h>
int main()
{
char c;
int chars = 0;
int numbers = 0;
int spaces = 0;
int others = 0;
printf("请输入一个字符串，输入完成后敲击回车:");
while(1)
{
c = getchar();
if(c=='\n')
{
break;
}
else
{
if(c>='0' && c<='9')
{
numbers++;
}
else if((c>='a' && c<='z')||(c>='A' && c<='Z'))
{
chars++;
}
else if(c == ' ')
{
spaces++;
}
else
{
others++;
}
}
}
printf("char=%d\tnumber=%d\tspace=%d\tother=%d\n", chars, numbers, spaces, others);
return 0;
}
C语言题目10
//求两个数字的最大公约数和最小公倍数
#include"stdio.h"
int main()
{
int min;
int max;
int a;
int b;
int swap;
int remain = 1;
printf("请输入数字a:");
scanf("%d", &max);
printf("请输入数字b:");
scanf("%d", &min);
a = max;
b = min;
if(min > max)
{
swap = min;
min = max;
max = swap;
}
while(remain!=0)
{
remain = max%min;
max = min;
min = remain;
}
printf("最大公约数为:%d\n", max);
printf("最小公倍数为:%d\n", a*b/max);
return 1;
}
C语言题目9
//打印100以内的素数
#include"stdio.h"
int main()
{
int i;
int j;
int flag = 1;
for(i=2; i<100; i++)
{
flag = 1;
for(j=i-1; j>1; j--)
{
if(i%j == 0)
{
flag = 0;
break;
}
}
if(flag == 1)
{
printf("%d ", i);
}
}
printf("\n");
return 1;
}
C语言题目8
//有两个集合A={a,b,f,d,e} B={g,b,h,d,e,e} 合并为一个集合C， C中的元素是递增排列的，并且无重复元素。
//这道题也可以拼接两个集合到一个集合，然后排序消重。
#include"stdio.h"
int getmin(char k[], int n)
{
int min = -1;
int i;
for(i=0; i<n; i++)
{
if(k[i] != '#')
{
min = i;
}
}
if(min != -1)
{
for(i=0; i<5; i++)
{
if((k[i] == '#') || i == min)
{
continue;
}
if(k[i] < k[min])
{
min = i;
}
else if(k[i] == k[min])
{
k[i] = '#';
}
}
}
return min;
}
int main()
{
char a[5] = {'a','b','f','d','e'};
char b[6] = {'g','b','h','d','e','e'};
char c[12] = {};
int i = 0;
int j = 0;
int mina;
int minb;
int k = 0;
int flaga = 0;
int flagb = 0;
while(1)
{
mina = getmin(a,5);
minb = getmin(b,6);
if((mina != -1) && (minb != -1))
{
if(a[mina] < b[minb])
{
c[k] = a[mina];
a[mina] = '#';
k = k+1;
}
else if(a[mina] == b[minb])
{
c[k] = a[mina];
a[mina] = '#';
b[minb] = '#';
k = k+1;
}
else
{
c[k] = b[minb];
b[minb] = '#';
k = k+1;
}
}
else if((mina == -1) && (minb != -1))
{
c[k] = b[minb];
b[minb] = '#';
k = k+1;
}
else if((mina != -1) && (minb == -1))
{
c[k] = a[mina];
a[mina] = '#';
k = k+1;
}
else
{
c[k] = '#';
break;
}
}
i = 0;
while(i<12)
{
if(c[i] == '#')
{
printf("\n");
break;
}
else
{
printf("%c ", c[i]);
i = i+1;
}
}
}
C语言题目7
//已知n=0, f(n)=0; n=1, f(n)=1; n>=2, f(n)=[f(n-1)+1]*[f(n-2)-1], 求f(50)。
//注意：double也只能存储到f(18)。这道题应该不会是要考大整数乘法，所以题目出的有问题。
//这道题也可以用递归，但是递归运行速度比较慢，因为有大量重复计算，所以可以按照我的方法用链表。
//如果真的想要求出f(50)，可以用链表存储两个计算因子和计算结果，转化成大整数加法，来求大整数乘法。
//把乘法转化成加法计算，比如3×4 = 4+4+4
//以下是前20个数字的运行结果
//0:0.000000
//1:1.000000
//2:-2.000000
//3:-0.000000
//4:-3.000000
//5:2.000000
//6:-12.000000
//7:-11.000000
//8:130.000000
//9:-1572.000000
//10:-202659.000000
//11:318781034.000000
//12:-64604164553100.000000
//13:-20594582312338883346432.000000
//14:1330495784608724577563601617989664768.000000
//15:-27401004952344282189274505435890321300250290209030873808896.000000
//16:-36456921583136852485661222029923476042013772777779791392989314785761343035841829994866675810304.000000
//17:9989562888467599682490178454054145458467604867235829190066412851063708189685362766757643907501070026206
//55223767942258990500160700908909006358607921414144.000000
//18:-364188710874677328890688166883341655997440487840343568582579911102615655287833051471441667841754329590
//03497968188602575038253510532501707167150463676880377107100124867120157208280951966764835855418677107978
//345178504768013589268641567045778554224640.000000
//19:-inf
//20:inf
#include"stdio.h"
#include"stdlib.h"
typedef struct node{
double x;
struct node *next;
}llnode;
void generate(llnode *head, int n)
{
llnode *p;
llnode *m;
double fn2 = 0;
double fn1 = 0;
int i = 0;
if(n==1)
{
m = (llnode *)malloc(sizeof(llnode));
m->x = 1;
m->next = NULL;
head->next = m;
}
else
{
p = head;
while(i<n-2)
{
p = p->next;
i = i+1;
}
fn2 = p->x;
p = p->next;
fn1 = p->x;
m = (llnode *)malloc(sizeof(llnode));
m->x = (fn1+1)*(fn2-1);
m->next = NULL;
p->next = m;
}
}
double f(llnode* head, int n)
{
int i=0;
llnode *p = head;
while(i<n)
{
p = p->next;
i = i+1;
}
return p->x;
}
int main()
{
llnode *head;
int n = 0;
int last = 20;//要求的最大结果
double result=0;
head = (llnode *)malloc(sizeof(llnode));
head->x = 0;
head->next = NULL;
for(n=1; n<=last; n++)
{
generate(head, n);
}
for(n=0; n<=last; n++)
{
result = f(head, n);
printf("%d:", n);
printf("%lf\n", result);
}
return 1;
}