主题：那个好心的高手帮忙做一下这几条简单的程序设计题?
Gillian
[专家分：0] 发布于 2005-02-17 03:37:00
[em2][em3] 给些相关资料的网站连接给小女子都可以~感激不尽
死人变态老师竟然要我们做这样的题目~就要开学了~我还没做完555555[em10]
1、 运动会分数统计
任务：参加运动会有n个学校，学校编号为1……n。比赛分成m个男子项目，和w个女子项目。项目编号为男子1……m，女子m+1……m+w。不同的项目取前五名或前三名积分；取前五名的积分分别为：7、5、3、2、1，前三名的积分分别为：5、3、2；哪些取前五名或前三名由学生自己设定。（m<=20,n<=20）
功能要求：1).可以输入各个项目的前三名或前五名的成绩；
2)．能统计各学校总分，
3)．可以按学校编号、学校总分、男女团体总分排序输出；
4).可以按学校编号查询学校某个项目的情况；可以按项目编号查询取得前三或前五名的学校。
规定：输入数据形式和范围：20以内的整数（如果做得更好可以输入学校的名称，运动项目的名称）
输出形式：有中文提示，各学校分数为整形
界面要求：有合理的提示，每个功能可以设立菜单，根据提示，可以完成相关的功能要求。
存储结构：学生自己根据系统功能要求自己设计，但是要求运动会的相关数据要存储在数据文件中。（数据文件的数据读写方法等相关内容在c语言程序设计的书上，请自学解决）请在最后的上交资料中指明你用到的存储结构；
测试数据：要求使用1、全部合法数据；2、整体非法数据；3、局部非法数据。进行程序测试，以保证程序的稳定。测试数据及测试结果请在上交的资料中写明；
2、 一元多项式计算
任务：能够按照指数降序排列建立并输出多项式；
能够完成两个多项式的相加、相减，并将结果输入；
在上交资料中请写明：存储结构、多项式相加的基本过程的算法（可以使用程序流程图） 、源程序、测试数据和结果、算法的时间复杂度、另外可以提出算法的改进方法；
3、 猴子选大王
任务：一堆猴子都有编号，编号是1，2，3 ...m ,这群猴子（m个）按照1-m的顺序围坐一圈，从第1开始数，每数到第N个，该猴子就要离开此圈，这样依次下来，直到圈中只剩下最后一只猴子，则该猴子为大王。
要求：
输入数据：输入m,n m,n 为整数，n<m
输出形式：中文提示按照m个猴子，数n 个数的方法，输出为大王的猴子是几号 ，建立一个函数来实现此功能
4、最小生成树问题**
问题描述：若要在n个城市之间建设通信网络，只需要架设n-1条线路即可。
如何以最低的经济建设这个通信网，是一个网的最小生成树
基本要求：
(1)利用克鲁斯卡尔算法求网的最小生成树。
(2)以文本形式输出生成树中各条边以及他们的权值
5、图的建立及输出
任务：建立图的存储结构（图的类型可以是有向图、无向图、有向网、无向网，学生可以任选两种类型），能够输入图的顶点和边的信息，并存储到相应存储结构中，而后输出图的邻接矩阵。
6、拓扑排序
任务：编写函数实现图的拓扑排序。
请尽量写出每一步的算法说明~~~
回复列表 （共25个回复）
11 楼
kingwei [专家分：3960] 发布于 2005-02-21 21:29:00
帮你写了一下猴子.
#include <stdio.h>
#include <conio.h>
#define MAXM 1000
int workarr[MAXM];
int Work(int m, int n)
{
int i, j, k = 0;
for (i=0; i<m; i++)
workarr[i] = i+1;
for (i=m; i>1; i--)
{
k = (k+n-1)%i;
for (j=k+1; j<i; j++)
workarr[j-1] = workarr[j];
}
return workarr[0];
}
int main()
{
int m, n, res;
printf("输入m,n (m>n): ");
scanf("%d%d", &m, &n);
res = Work(m, n);
printf("按照%d个猴子，数%d个数的方法，大王的猴子是%d号\n", m, n, res);
getch();
return 0;
}
12 楼
mengxin0839 [专家分：0] 发布于 2005-02-22 17:12:00
高手呢？我也想知道运动会问题、一元多项式问题、图的输出这三个问题的C程序啊！
着急用呢！我们老师跟我们的题目要求这三个题目居然和楼主的一样。
13 楼
Gillian [专家分：0] 发布于 2005-02-22 21:04:00
十分感谢kingwei GG,还有那位高手能帮忙做一做剩下的那些题目啊? 小女感激不尽
14 楼
kingwei [专家分：3960] 发布于 2005-02-23 21:39:00
帮mengxin0839写的多项式,顺便贴一下,没测过
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
typedef struct PolyNode
{
int coef;
int expn;
struct PolyNode *next;
}PolyNode, *PolyType;
void FreePoly(PolyType *poly)
{
PolyNode *p = *poly, *q;
while (p != NULL)
{
q = p->next;
free(p);
p = q;
}
*poly = NULL;
}
void DispPoly(PolyType poly)
{
PolyNode *p = poly->next;
printf("%d", poly->expn);
while (p != NULL)
{
printf(", %d, %d", p->coef, p->expn);
p = p->next;
}
printf("\n");
}
void AddPoly(PolyType *res, PolyType left, PolyType right)
{
PolyNode *t, *p = left->next, *q = right->next;
*res = (PolyNode *)malloc(sizeof(PolyNode));
(*res)->expn = (*res)->coef = 0;
(*res)->next = NULL;
t = *res;
while (p!=NULL || q!=NULL)
{
if (p!=NULL && q!=NULL && p->expn==q->expn)
{
if (p->coef+q->coef != 0)
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = p->expn;
t->coef = p->coef+q->coef;
t->next = NULL;
(*res)->expn++;
}
p = p->next;
q = q->next;
}
else if (q==NULL || (p!=NULL && q!=NULL && p->expn>q->expn))
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = p->expn;
t->coef = p->coef;
t->next = NULL;
(*res)->expn++;
p = p->next;
}
else
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = q->expn;
t->coef = q->coef;
t->next = NULL;
(*res)->expn++;
q = q->next;
}
}
}
void SubPoly(PolyType *res, PolyType left, PolyType right)
{
PolyNode *t, *p = left->next, *q = right->next;
*res = (PolyNode *)malloc(sizeof(PolyNode));
(*res)->expn = (*res)->coef = 0;
(*res)->next = NULL;
t = *res;
while (p!=NULL || q!=NULL)
{
if (p!=NULL && q!=NULL && p->expn==q->expn)
{
if (p->coef-q->coef != 0)
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = p->expn;
t->coef = p->coef-q->coef;
t->next = NULL;
(*res)->expn++;
}
p = p->next;
q = q->next;
}
else if (q==NULL || (p!=NULL && q!=NULL && p->expn>q->expn))
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = p->expn;
t->coef = p->coef;
t->next = NULL;
(*res)->expn++;
p = p->next;
}
else
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = q->expn;
t->coef = -q->coef;
t->next = NULL;
(*res)->expn++;
q = q->next;
}
}
}
void ReadPoly(PolyType *poly)
{
PolyNode *p, *q;
int coef, expn;
*poly = (PolyNode *)malloc(sizeof(PolyNode));
(*poly)->expn = (*poly)->coef = 0;
(*poly)->next = NULL;
scanf("%d%d", &coef, &expn);
while (expn >= 0)
{
p = *poly;
q = (*poly)->next;
while (q!=NULL && q->expn>expn)
{
p = p->next;
q = q->next;
}
if (q!=NULL && q->expn==expn)
{
if (q->coef+coef == 0)
{
p->next = q->next;
((*poly)->expn)--;
free(q);
}
else
q->coef += coef;
}
else
{
p->next = (PolyNode *)malloc(sizeof(PolyNode));
p = p->next;
p->expn = expn;
p->coef = coef;
p->next = q;
((*poly)->expn)++;
}
scanf("%d%d", &coef, &expn);
}
}
int main()
{
PolyType left, right, res;
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
ReadPoly(&left);
ReadPoly(&right);
printf("polynomial #1: ");
DispPoly(left);
printf("polynomial #2: ");
DispPoly(right);
AddPoly(&res, left, right);
printf("polynomial #1 + #2: ");
DispPoly(res);
FreePoly(&res);
SubPoly(&res, left, right);
printf("polynomial #1 - #2: ");
DispPoly(res);
FreePoly(&res);
FreePoly(&left);
FreePoly(&right);
getch();
return 0;
}
15 楼
kingwei [专家分：3960] 发布于 2005-02-23 21:41:00
输入样例
1 1
2 3
4 1
5 6
-2 3
-99 44
-1 -1
3 5
99 44
4 6
-5 6
3 4
4 6
7 7
-1 0
-1 -1
输出
polynomial #1: 3, -99, 44, 5, 6, 5, 1
polynomial #2: 6, 99, 44, 7, 7, 3, 6, 3, 5, 3, 4, -1, 0
polynomial #1 + #2: 6, 7, 7, 8, 6, 3, 5, 3, 4, 5, 1, -1, 0
polynomial #1 - #2: 7, -198, 44, -7, 7, 2, 6, -3, 5, -3, 4, 5, 1, 1, 0
16 楼
kingwei [专家分：3960] 发布于 2005-02-24 00:20:00
应mengxin0839的要求改成屏幕提示的了.又加了两个功能,求导和求值.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <conio.h>
typedef struct PolyNode
{
int coef;
int expn;
struct PolyNode *next;
}PolyNode, *PolyType;
void FreePoly(PolyType *poly)
{
PolyNode *p = *poly, *q;
while (p != NULL)
{
q = p->next;
free(p);
p = q;
}
*poly = NULL;
}
void DispPoly(PolyType poly)
{
PolyNode *p = poly->next;
if (p == NULL)
{
printf("0\n");
return;
}
while (p != NULL)
{
if (p->expn > 0)
printf("%+dx^%d", p->coef, p->expn);
else
printf("%+d", p->coef);
p = p->next;
}
printf("\n");
}
void AddPoly(PolyType *res, PolyType left, PolyType right)
{
PolyNode *t, *p = left->next, *q = right->next;
*res = (PolyNode *)malloc(sizeof(PolyNode));
(*res)->expn = (*res)->coef = 0;
(*res)->next = NULL;
t = *res;
while (p!=NULL || q!=NULL)
{
if (p!=NULL && q!=NULL && p->expn==q->expn)
{
if (p->coef+q->coef != 0)
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = p->expn;
t->coef = p->coef+q->coef;
t->next = NULL;
(*res)->expn++;
}
p = p->next;
q = q->next;
}
else if (q==NULL || (p!=NULL && q!=NULL && p->expn>q->expn))
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = p->expn;
t->coef = p->coef;
t->next = NULL;
(*res)->expn++;
p = p->next;
}
else
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = q->expn;
t->coef = q->coef;
t->next = NULL;
(*res)->expn++;
q = q->next;
}
}
}
void SubPoly(PolyType *res, PolyType left, PolyType right)
{
PolyNode *t, *p = left->next, *q = right->next;
*res = (PolyNode *)malloc(sizeof(PolyNode));
(*res)->expn = (*res)->coef = 0;
(*res)->next = NULL;
t = *res;
while (p!=NULL || q!=NULL)
{
if (p!=NULL && q!=NULL && p->expn==q->expn)
{
if (p->coef-q->coef != 0)
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = p->expn;
t->coef = p->coef-q->coef;
t->next = NULL;
(*res)->expn++;
}
p = p->next;
q = q->next;
}
else if (q==NULL || (p!=NULL && q!=NULL && p->expn>q->expn))
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = p->expn;
t->coef = p->coef;
t->next = NULL;
(*res)->expn++;
p = p->next;
}
else
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->expn = q->expn;
t->coef = -q->coef;
t->next = NULL;
(*res)->expn++;
q = q->next;
}
}
}
void DifPoly(PolyType *res, PolyType poly)
{
PolyNode *p = poly->next, *t;
*res = (PolyNode *)malloc(sizeof(PolyNode));
(*res)->expn = (*res)->coef = 0;
(*res)->next = NULL;
t = *res;
while (p != NULL)
{
if (p->expn > 0)
{
t->next = (PolyNode *)malloc(sizeof(PolyNode));
t = t->next;
t->coef = p->coef*p->expn;
t->expn = p->expn-1;
t->next = NULL;
(*res)->expn++;
}
p = p->next;
}
}
long CalcPoly(PolyType poly, int x)
{
long res = 0;
PolyNode *p = poly->next;
while (p != NULL)
{
res += p->coef*(long)pow(x, p->expn);
p = p->next;
}
return res;
}
void ReadPoly(PolyType *poly)
{
PolyNode *p, *q;
int coef, expn;
*poly = (PolyNode *)malloc(sizeof(PolyNode));
(*poly)->expn = (*poly)->coef = 0;
(*poly)->next = NULL;
printf("input coef: ");
scanf("%d", &coef);
printf("input exp: ");
scanf("%d", &expn);
while (!(expn==0 && coef==0))
{
p = *poly;
q = (*poly)->next;
while (q!=NULL && q->expn>expn)
{
p = p->next;
q = q->next;
}
if (q!=NULL && q->expn==expn)
{
if (q->coef+coef == 0)
{
p->next = q->next;
((*poly)->expn)--;
free(q);
}
else
q->coef += coef;
}
else if (coef != 0)
{
p->next = (PolyNode *)malloc(sizeof(PolyNode));
p = p->next;
p->expn = expn;
p->coef = coef;
p->next = q;
((*poly)->expn)++;
}
printf("input coef: ");
scanf("%d", &coef);
printf("input exp: ");
scanf("%d", &expn);
}
}
17 楼
kingwei [专家分：3960] 发布于 2005-02-24 00:20:00
void DispUsage()
{
printf("\n1:add\n");
printf("2:sub\n");
printf("3:mul (NOT SUPPORTED)\n");
printf("4:qiu_dao\n");
printf("5:data in x\n");
printf("6:exit\n\n");
}
int main()
{
PolyType left, right, res;
int op, x;
long val;
DispUsage();
scanf("%d", &op);
while (op != 6)
{
switch (op)
{
case 1: printf("input the first\n"); ReadPoly(&left);
printf("input the second\n"); ReadPoly(&right);
AddPoly(&res, left, right);
printf("the first: "); DispPoly(left);
printf("the second: "); DispPoly(right);
printf("sum(+) is: "); DispPoly(res);
FreePoly(&left); FreePoly(&right); FreePoly(&res);
break;
case 2: printf("input the first\n"); ReadPoly(&left);
printf("input the second\n"); ReadPoly(&right);
SubPoly(&res, left, right);
printf("the first: "); DispPoly(left);
printf("the second: "); DispPoly(right);
printf("sum(-) is: "); DispPoly(res);
FreePoly(&left); FreePoly(&right); FreePoly(&res);
break;
case 4: ReadPoly(&left);
DifPoly(&res, left);
printf("before qiudao: "); DispPoly(left);
printf("after qiudao: "); DispPoly(res);
FreePoly(&left); FreePoly(&res);
break;
case 5: ReadPoly(&left);
printf("input x: ");
scanf("%d", &x);
val = CalcPoly(left, x);
DispPoly(left);
printf("x = %d\n", x);
printf("sum(x) is: %ld\n", val);
FreePoly(&left);
break;
default:DispUsage();
}
scanf("%d", &op);
}
return 0;
}
18 楼
7645690 [专家分：0] 发布于 2006-06-23 13:54:00
怎么没有第5题“图的建立及输出”呢？哪位高手提供以下呀，很急！
19 楼
7645690 [专家分：0] 发布于 2006-06-23 14:38:00
有第5题的程序吗？
建立图的存储结构（图的类型可以是有向图、无向图、有向网、无向网，学生可以任选两种类型），能够输入图的顶点和边的信息，并存储到相应存储结构中，而后输出图的邻接矩阵。
谢谢~~有的发一份给我好么?
jianjin_2002@163.com
20 楼
7645690 [专家分：0] 发布于 2006-06-23 14:41:00
有第5题的程序吗？
建立图的存储结构（图的类型可以是有向图、无向图、有向网、无向网，学生可以任选两种类型），能够输入图的顶点和边的信息，并存储到相应存储结构中，而后输出图的邻接矩阵。
这样的程序么???谢谢~~有的发一份给我好么?
jianjin_2002@163.com
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