Problem Description
There will be n schools participating in the sports, and the school numbers are 1~n. The competition is divided into m men's events and w women's events. The item numbers are 1~m for men and m+1~m+w for women. Different items take the top five or top three points. The points for the top five are: 7, 5, 3, 2, 1; the points for the top three are: 5, 3, 2. Which items take the top five and which items take the top three can be set by yourself. (m<=20, n<=20)
functional requirements
(1) You can enter the top three or top five results of each item;
(2) Can count the total scores of each school;
(3) It can be sorted and output by school number or name, school total score, and total score of male and female groups;
(4) You can query the situation of a certain project of a school according to the school number; you can query the top three or top five schools according to the project number.
(5) The data is stored in files and can be queried at any time.
(6) Regulations
- Input data form and range: You can input the name of the school and the name of the sports event.
- Output form: There are Chinese prompts, and the scores of each school are integers.
- Interface requirements: There are reasonable prompts, and menus can be set up for each function, and relevant functional requirements can be completed according to the prompts.
- Storage structure: According to the requirements of the system function, the attack is involved, but the relevant data of the sports meeting is required to be stored in the file.
- Test data: Use 1. All legal data; 2. Overall illegal data; 3. Partial illegal data. Test to ensure the stability of the program.
Above code:
#include <stdio.h>
#include <stdlib.h>
#define NAME 17 //字符数
#define NUMU 20
#define NUMP 20
#define H 5 //每一行输出的学校或项目数
FILE *fp;
const char wenjian[] = "system.txt"; //文件命名
typedef struct link
{
int gram[3][NUMP];//gram[0]存男生项目，gram[1]存女生项目,gram[2]存这个学校中男女在此项目上的总分
int sum[3];//sum[0]存男生总得分，sum[1]存女生总得分，sum[2]存学校总得分
}link;
typedef struct MESSAGE
{
int b[4]; // b[0-3]依次为(学校个数，项目个数，男生项目个数，女生项目个数
char xuexiao[NUMU][NAME];//学校名称
int sex[NUMU];//性别（区分男女子项目）
struct link spot[NUMU];//添加积分情况
int junction[NUMP];//决定取前三或前五
int count[NUMP][5];//项目的学校编号
char xiangmu[NUMP][NAME];//项目名称
} MESSAGE;
int junction[][5] = {
{5, 3, 2, 0, 0}, {7, 5, 3, 2, 1}};
char sex[][3] = {"男", "女"};
char fenlei[][5] = {"男子", "女子", "学校"};
int gainint(int *p, int a, int b);//输入数据p，输入数据下限a，输入数据上限b
int gainchar(char A[], int min, int max);//输入字符A，输入字符数量下限min，输入字符数量上限max
int bijiao(char a[], char b[]);//等待进行匹配查找的字符串a，进行匹配查找的关键字b
void record(char school[][NAME], char game[][NAME], int b[], MESSAGE *temp);
int xianshi();
int jianyan(char b[], int T);//检验字符是否符合要求,T=0检验当前字符串是否全为汉字 T=1检验学校,T=2检验比赛
int jianyan2(char b[], int T);
void paixu(MESSAGE *temp, int select,FILE *fp);
char key[][3] = {"！", "。", "？", "《", "》", "，", "｛", "｝", "（", "）", "￥", "‘", "’", "：", "；", "—", "”", "“", "、", "·", "~", "\0"};
char school[][5] = {"大学", "学院", "华侨", "校区", "政法", "管理", "经济", "科学", "理工", "科技", "医科", "医药", "财经", "技术", "职业", "师范", "中国", "海洋", "石油", "民族", "外贸", "分校", "建筑", "工业", "外语", "\0"};
char program[][3] = {"跳", "滑", "冰", "射", "骑", "击", "跑", "拉", "道", "球", "举", "米", "水", "摔", "爬", "泳", "拔", "车", "踩", "接", "划", "体", "田", "人", "赛", "\0"};
int read(MESSAGE *message,FILE *fp)
{
if ((fp = fopen(wenjian, "r")) == NULL)
{
if ((fp = fopen(wenjian, "w+")) != NULL)
{
fclose(fp);
if ((fp = fopen(wenjian, "r")) == NULL)
return printf("文件读取失败!\n");
}
}
fread(message, sizeof(MESSAGE), 1, fp);
fclose(fp);
return 0;
}
int main()
{
MESSAGE SG = {0}, *message = &SG;
int i, j, k, select, choice, now, next;
if (read(message,fp))
{
printf("请按任意键进行下一步\n");
getch();
}
do
{
system("cls");
select = xianshi();
if (select == 1)
record(message->xuexiao, message->xiangmu, message->b, message);
else if (message->b[0])
{
if (select >= 2 && select <= 4)
{
fp = fopen(wenjian, "a+");
paixu(message, select - 2,fp);
fclose(fp);
}
else if (select == 5)
{
fp = fopen(wenjian, "a+");
for (k = 0; k < message->b[0]; k++) //输出学校名称选项 每五个一行
printf("%d:%-13s%c", k + 1, message->xuexiao[k], (k + 1) % H ? ' ' : '\n');
printf("\n输入学校前的编号[1,%d]:", message->b[0]);
choice = gainint(&choice, 1, message->b[0]) - 1;
next = 0;
while (next < message->b[1])
{
now = next;
if ((next + H) < (message->b[1]))
next = next + H;
else
next = message->b[1];
for (i = now; i < next; i++)
{
fprintf(fp,"(%s-%d)%-8s ", sex[message->sex[i]], message->junction[i], message->xiangmu[i]);
printf("(%s-%d)%-8s ", sex[message->sex[i]], message->junction[i], message->xiangmu[i]);
} //输出项目信息
fprintf(fp,"\n");
printf("\n");
for (j = 0; j < 3; j++)
{
fprintf(fp,"%-s: ", fenlei[j]);
printf("%-s: ", fenlei[j]);
for (k = now; k < next; k++)
{
if (j == message->sex[k] || j == 2)
{
fprintf(fp,"%02d分 \t", (message->spot + choice)->gram[j][k]);
printf("%02d分 \t", (message->spot + choice)->gram[j][k]);
}
else
{
fprintf(fp,"****** \t");
printf("****** \t");
}
}
fprintf(fp,"\n");
printf("\n");
}
fprintf(fp,"\n");
printf("\n");
}
fclose(fp);
}
else if (select == 6)
{
fp = fopen(wenjian, "a+");
for (i = 0; i < message->b[1]; i++)
{
printf("\t\t第%d个比赛项目: %-13s<%s子> 只录入前%d名\n", i + 1, message->xiangmu[i], sex[message->sex[i]], message->junction[i]);
fprintf(fp,"\t\t第%d个比赛项目: %-13s<%s子> 只录入前%d名\n", i + 1, message->xiangmu[i], sex[message->sex[i]], message->junction[i]);
for (j = 0; j < message->junction[i]; j++)
{
fprintf(fp,"\t第%d名:%-13s%02d分", j + 1, message->xuexiao[message->count[i][j]], junction[(message->junction[i]) / 4][j]);
printf("\t第%d名:%-13s%02d分", j + 1, message->xuexiao[message->count[i][j]], junction[(message->junction[i]) / 4][j]);
}
printf("\n\n");
fprintf(fp,"\n\n");
}
fclose(fp);
}
else if (select == 9)
{
printf("确认清空请按1 否则请按0:");
if (gainint(&choice, 0, 1))
for (k = 0; k < NUMP; k++) //对申请出来的结构体中的所有项目初始化
for (i = 0; i < 3; i++)
{
(message->spot + k)->sum[i] = 0; // sum[]初始化
for (j = 0; j < NUMP; j++)
(message->spot + k)->gram[i][j] = 0; //项目初始化
message->b[i] = 0;
}
/*if (write(message,fp))
return 0;*/
read(message,fp);
}
else if(select==7)
{
system(wenjian);
}
}
printf("\n请按任意键继续…\n");
getch();
} while (select != 8);
return 0;
}
int bijiao(char a[], char b[])
{
int i = 0, j = 0;
while (a[i] && b[j])
if (a[i++] == b[j])
{
j++;
}
else
{
i = i - j;
j = 0;
}
if (b[j] == '\0')
return i - j + 1;//若匹配成功则返回第一次匹配成功的下标
else
return 0;
}
int gainint(int *p, int a, int b)
{
do
{
*p = 0;
scanf("%d", p);
while (getchar() != '\n')
;
if (*p > b || *p < a)
printf("输入有误,请重新输入[%d--%d]:", a, b);
} while (*p > b || *p < a);
return *p;//返回输入数据
}
int gainchar(char A[], int min, int max)
{
int B, C;
do
{
A[max] = B = C = 0;
while ((A[B++] = getchar()) != '\n' && B < max)
;
if (A[B - 1] != '\n')
while (getchar() != '\n' && ++C)
;//如果输入的最后一个字符不是回车，那么将剩余长度计到C
else
A[--B] = 0;//否则把录进数组的回车删掉，即为成功输入字符。
if (C || B && B < min)
printf("录入的字符串长度为%d字节\n只录入(%d--%d)个字符!\n", B + C, min, max);
} while (C || B < min);
return B;//返回录入成功字符数量
}
int jianyan(char b[], int T) //检验b[]是否符合实际
{
int d = 0, i, LEN = -1, F;
while (b[++LEN])
;
for (i = 0; i < LEN; i++)
if (b[i] > 0) //汉字ascll编码每一个字节为负数
return printf("输入中含有非中文字符!");
for (i = 0; key[i][0]; i++)
{
F = bijiao(b, key[i]); //匹配中文标点符号
if (F && F % 2)
return printf("输入中含有非汉字中文字符!");
}
if (T == 1) //检验学校的名称
{
for (i = 0; school[i][0]; i++)
{
F = bijiao(b, school[i]);
if (F && F % 2)
d++;
}
if (!d)
return printf("查无此校~学校名称输入可能存在错误!");
}
if (T == 2) //检验比赛的名称
{
for (i = 0; program[i][0]; i++)
{
F = bijiao(b, program[i]);
if (F && F % 2)
d++;
}
if (!d)
return printf("没有发现相关比赛~项目名称输入可能存在错误!");
}
return 0;
}
int jianyan2(char b[], int T) //检验b[]是否符合实际
{
int d = 0, i, LEN = -1, F;
if (T == 2) //检验比赛的名称
{
for (i = 0; program[i][0]; i++)
{
F = bijiao(b, program[i]);
if (F )
d++;
}
if (!d)
return printf("没有发现相关比赛~项目名称输入可能存在错误!");
}
return 0;
}
void record(char school[][NAME], char game[][NAME], int b[], MESSAGE *temp) //记录数据；school为学校名，game为项目名，b（参考结构体中定义），temp信息结构体
{
char choose[][13] = {"录入学校数目", "总计项目数目", "男子项目数目"}; //提示用户输入内容
int scope[][3] = {
{5, 1, 0}, {NUMU, NUMP, NUMU}}; //输入的各个内容的范围
int i, j, k, pan, count, choice, man;
scope[1][0] = NUMU - b[0]; //剩余录入的学校的数量（主要为后续加入数据设计）
scope[1][1] = NUMP - b[1]; //剩余录入的项目的数量（主要为后续加入数据设计）
if (b[0])
{
scope[0][0] = scope[0][1] = 0;
}//当用户已经进入此函数输入过一次后,改变条件将学校个数项目个数下限改为0
for (i = 0; i < 3; i++)
{
printf("\t%s(%d--%d)\n\t:", choose[i], scope[0][i], scope[1][i]);
gainint(&b[i], scope[0][i], scope[1][i]);
if (i == 1)
{
scope[1][2] = b[1]; //输入项目完毕后，改变男生项目的长度
if (!b[0] && !b[1])//输入为零时进行下一步
{
b[2] = b[1];
i++;
}
}
if (i == 2)
{
if (!b[0] || !b[1])
{
return;
}
b[3] = b[1] - b[2]; //男生输入完毕时，女生的项目=总项目-男生项目
}
}
for (i = NUMU - scope[1][0]; i < b[0] + NUMU - scope[1][0]; i++)
do
{
k = 0;
printf("请输入第%d个学校的名称(只录入%d-%d字符):\n", i + 1, 8, NAME);
gainchar(school[i], 8, NAME);
for (j = 0; j < i; j++) //对输入的学校名称进行查重
if (bijiao(school[j], school[i]) && bijiao(school[i], school[j])) //之前的学校和当前录入的学校重复
k++;
if (k)
printf("该学校已经录入!");
} while (jianyan(school[i], 1) || k);
b[0] += NUMU - scope[1][0]; //计算出目前的学校总数（主要为后续加入数据设计）
for (i = NUMP - scope[1][1]; i < b[1] + NUMP - scope[1][1]; i++) //录入项目
{
temp->sex[i] = (b[1] + NUMP - scope[1][1] - i) <= b[3]; //判断性别,0为男,1为女，先输入男
do
{
k = 0;
printf("请输入第%d个项目(%s)名称(只录入%d-%d字符):\n", i + 1, sex[temp->sex[i]], 4, NAME); //输入项目名称
gainchar(game[i], 4, NAME);
for (j = 0; j < i; j++)
if (bijiao(game[j], game[i]))
k++;
if (k)
printf("项目已录入成功!");
pan = (k || jianyan2(game[i], 2)); //区分分录入前三名还是前五名
if (!pan) //当输入满足要求时 继续询问项目的录入名次
do
{
printf("该项目录入成绩: 3:取前三 5:取前五:"); //询问录入的时前几名
gainint(&(temp->junction[i]), 3, 5); //将第i+1个项目的分数情况放在头节点的junction[i]数组里
} while ((temp->junction[i]) == 4); //只录入3和5
} while (pan); //当输入不满足要求时重新输入
system("cls");
for (k = 0; k < b[0]; k++) //输出学校名称选项 每五个一行
printf("%d:%-13s%c", k + 1, school[k], (k + 1) % H ? ' ' : '\n');
printf("\n\n获得 %s:(%s子)的前%d名(填学校前的编号):\n", game[i], sex[temp->sex[i]], temp->junction[i]);
for (j = 0; j < temp->junction[i]; j++) //每次录入的项目个数=temp->junction[i]
{
do
{
pan = 0;
printf("第%d名(%d-%d):", j + 1, 1, b[0]);
count = gainint(&count, 1, b[0]) - 1; //范围1-b[0]
(temp->count[i][j]) = count; // temp->count[i][j]中的数值就是第i个学校,名次为j
for (k = 0; k < j; k++) //检验名次是否录入重复
if (temp->count[i][k] == count)
pan = printf("一个学校只能获得一个名次!\n");
} while (pan);
choice = junction[(temp->junction[i]) / 4][j];
man = temp->sex[i];
(temp->spot + count)->gram[man][i] = choice; //把choice积分存入第count个学校的第i个项目的gram[man][i]里
(temp->spot + count)->gram[2][i] += choice; //修改第count个学校的该项目总分
(temp->spot + count)->sum[man] += choice; //修改第count个学校的男女团体总积分
(temp->spot + count)->sum[2] += choice; //修改学校的总积分
}
}
b[1] += NUMP - scope[1][1]; //修改录入的项目的总数量（主要为后续加入数据设计）
}
void paixu(MESSAGE *temp, int select,FILE *fp) //选择排序
{
int i, j, k, m, a, b, n, school[2][NUMU] = {0};
n = temp->b[0];
printf("升序请按0 降序请按1:");
gainint(&b, 0, 1);
for (i = 0; i < n; i++)
{
school[1][i] = i; //记录学校当前的位置编号
school[0][i] = (temp->spot + i)->sum[select];
}
printf("%s排名如下\t（以第%d列分数为准）:\n\n", fenlei[select], select + 1);
fprintf(fp,"%s排名如下\t（以第%d列分数为准）:\n\n", fenlei[select], select + 1);
for (i = 0; i < n - 1; i++)
{
k = i;
for (j = i + 1; j < n; j++)
{
if (b && school[0][j] > school[0][k])
k = j;
else if (!b && school[0][j] < school[0][k])
k = j;
}
if (k != i)
for (m = 0; m < 2; m++)
{
a = school[m][k];
school[m][k] = school[m][i];
school[m][i] = a;
}
}
printf("名次: 学校名称 男子 女子 学校总分\n");
fprintf(fp,"名次: 学校名称 男子 女子 学校总分\n");
for (i = 0; i < n; i++)
{
k = (b ? i + 1 : n - i); //计算排名的名次显示方向
j = school[1][i]; // 记录学校的编号
printf("第%02d:%-13s%02d分\t%02d分\t%02d分\n", k, temp->xuexiao[j], (temp->spot + j)->sum[0], (temp->spot + j)->sum[1], (temp->spot + j)->sum[2]);
fprintf(fp,"第%02d:%-13s%02d分\t%02d分\t%02d分\n", k, temp->xuexiao[j], (temp->spot + j)->sum[0], (temp->spot + j)->sum[1], (temp->spot + j)->sum[2]);
}
}
int xianshi()
{
int a;
system("mode con cols=130 lines=30"); //控制台 宽度130 高度30
printf("\t\t\t\t运动会分数统计\n");
printf("* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *\n");
printf("\t* 功 能 表 积 分 规 则\n");
printf("\t* 1.添加学校及项目 *取前三名(1,2,3)\n");
printf("\t* 2.按男子团体总分排序输出 *积分分别为--5,3,2\n");
printf("\t* 3.按女子团体总分排序输出 \n");
printf("\t* 4.按学校总分排序输出 *取前五名(1,2,3,4,5)\n");
printf("\t* 5.按编号或名称查询学校情况 *积分分别为--7,5,3,2,1\n");
printf("\t* 6.显示比赛项目各学校名次情况 \n");
printf("\t* 7.读取文件 \n");
printf("\t* 8.保存信息并退出 \n");
printf("\t* 9.清空并恢复初始化界面 \n");
printf("\n");
printf("\t*注意：一个汉字占两个字符！！！\n");
printf("\t*注意：功能1可分多次选择！！！\n");
printf("\t*注意：录入项目时默认先男后女！！！\n");
printf("\t*注意：初次使用请先使用功能1录入数据！！！\n");
printf("* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *\n\t");
printf("请输入您的选择(1-9):");
return gainint(&a, 1, 9);
}interface:
file output:
