C/C++教程
UVA12412 A Typical Homework (a.k.a 师兄帮帮忙) 题解
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题意
实现一个学生信息管理系统,完成要求功能,并按要求输出.具体内容可以看[UVA12412](UVA12412 A Typical Homework (a.k.a 师兄帮帮忙) - 洛谷 | 计算机科学教育新生态 (luogu.com.cn))
输入输出样例
输入 #1
1 0011223344 1 John 79 98 91 100 0022334455 1 Tom 59 72 60 81 0011223344 2 Alice 100 100 100 100 2423475629 2 John 60 80 30 99 0 3 0022334455 John 0 5 1 2 0011223344 0 5 0 4 0
输出 #1
Welcome to Student Performance Management System (SPMS). 1 - Add 2 - Remove 3 - Query 4 - Show ranking 5 - Show Statistics 0 - Exit Please enter the SID, CID, name and four scores. Enter 0 to finish. Please enter the SID, CID, name and four scores. Enter 0 to finish. Please enter the SID, CID, name and four scores. Enter 0 to finish. Duplicated SID. Please enter the SID, CID, name and four scores. Enter 0 to finish. Please enter the SID, CID, name and four scores. Enter 0 to finish. Welcome to Student Performance Management System (SPMS). 1 - Add 2 - Remove 3 - Query 4 - Show ranking 5 - Show Statistics 0 - Exit Please enter SID or name. Enter 0 to finish. 2 0022334455 1 Tom 59 72 60 81 272 68.00 Please enter SID or name. Enter 0 to finish. 1 0011223344 1 John 79 98 91 100 368 92.00 3 2423475629 2 John 60 80 30 99 269 67.25 Please enter SID or name. Enter 0 to finish. Welcome to Student Performance Management System (SPMS). 1 - Add 2 - Remove 3 - Query 4 - Show ranking 5 - Show Statistics 0 - Exit Please enter class ID, 0 for the whole statistics. Chinese Average Score: 69.00 Number of passed students: 1 Number of failed students: 1 Mathematics Average Score: 85.00 Number of passed students: 2 Number of failed students: 0 English Average Score: 75.50 Number of passed students: 2 Number of failed students: 0 Programming Average Score: 90.50 Number of passed students: 2 Number of failed students: 0 Overall: Number of students who passed all subjects: 1 Number of students who passed 3 or more subjects: 2 Number of students who passed 2 or more subjects: 2 Number of students who passed 1 or more subjects: 2 Number of students who failed all subjects: 0 Welcome to Student Performance Management System (SPMS). 1 - Add 2 - Remove 3 - Query 4 - Show ranking 5 - Show Statistics 0 - Exit Please enter SID or name. Enter 0 to finish. 1 student(s) removed. Please enter SID or name. Enter 0 to finish. Welcome to Student Performance Management System (SPMS). 1 - Add 2 - Remove 3 - Query 4 - Show ranking 5 - Show Statistics 0 - Exit Please enter class ID, 0 for the whole statistics. Chinese Average Score: 59.50 Number of passed students: 1 Number of failed students: 1 Mathematics Average Score: 76.00 Number of passed students: 2 Number of failed students: 0 English Average Score: 45.00 Number of passed students: 1 Number of failed students: 1 Programming Average Score: 90.00 Number of passed students: 2 Number of failed students: 0 Overall: Number of students who passed all subjects: 0 Number of students who passed 3 or more subjects: 2 Number of students who passed 2 or more subjects: 2 Number of students who passed 1 or more subjects: 2 Number of students who failed all subjects: 0 Welcome to Student Performance Management System (SPMS). 1 - Add 2 - Remove 3 - Query 4 - Show ranking 5 - Show Statistics 0 - Exit Showing the ranklist hurts students’ self-esteem. Don’t do that. Welcome to Student Performance Management System (SPMS). 1 - Add 2 - Remove 3 - Query 4 - Show ranking 5 - Show Statistics 0 - Exit
思路
这道题和平时的编程大作业没有太大区别,没有什么思维难度,但格式要求非常严格,稍不注意就WA,建议使用自顶向下的方法来实现.参考紫书答案和洛谷题解.
代码
#include <stdio.h>
#include <string.h>
#define maxn 1005
#define maxl 1005
#define EPS 1e-5
// 定义结构体
struct stu {
char SID[maxn];
char name[maxn];
int CID;
int Ch, Ma, En, Pr, All;
// 无视这两个,写的时候没有用到
double Ave;
int rank;
};
stu st[maxl];
int n = 0;
int removed[maxl] = {0};
// 查重
int check(char* t) {
for (int i = 0; i < n; i++) {
if (!removed[i]) {
if (strcmp(t, st[i].SID) == 0) {
return 0;
}
}
}
return 1;
}
// 增加功能
void add() {
while (1) {
printf("Please enter the SID, CID, name and four scores. Enter 0 to finish.\n");
scanf("%s", st[n].SID);
if (strcmp(st[n].SID,"0" ) == 0) {
return;
}
scanf("%d%s%d%d%d%d", &st[n].CID, st[n].name, &st[n].Ch, &st[n].Ma, &st[n].En, &st[n].Pr);
if (check(st[n].SID) == 0) {
printf("Duplicated SID.\n");
continue;
} else {
st[n].All = st[n].Ch + st[n].En + st[n].Ma + st[n].Pr;
n++;
}
}
}
// 排名就是,有多少人比当前学生高分,这个学生的排名即人数加1
int rank(int s) {
int sum = 1;
for (int i = 0; i < n; i++) {
if (removed[i] == 1) {
continue;
}
if (st[i].All > s) {
sum++;
}
}
return sum;
}
// 查询,删除都要先查到学生,可以整合成一个函数
void RQ(int flag) {
while (1) {
printf("Please enter SID or name. Enter 0 to finish.\n");
char temp[maxn];
int r = 0;
scanf("%s", temp);
if (strcmp(temp, "0") == 0) {
return;
}
for (int i = 0; i < n; i++) {
if (removed[i] == 1) {
continue;
}
if (strcmp(temp, st[i].SID) == 0 || strcmp(temp, st[i].name) == 0) {
if (flag == 0) {
// 注意浮点数输出加一个EPS,缓解精度误差带来的问题
printf("%d %s %d %s %d %d %d %d %d %.2f\n", rank(st[i].All), st[i].SID, st[i].CID, st[i].name, st[i].Ch, st[i].Ma, st[i].En, st[i].Pr, st[i].All, st[i].All / 4.0 + EPS);
} else {
removed[i] = 1;
r++;
}
}
}
if (flag) {
printf("%d student(s) removed.\n", r);
}
}
}
void ST() {
printf("Please enter class ID, 0 for the whole statistics.\n");
int cid;
int chnum = 0, manum = 0, ennum = 0, prnum = 0;
int pass0 = 0, pass1 = 0, pass2 = 0, pass3 = 0, pass4 = 0;
int num = 0, passnum = 0;
int all = 0;
double cntot = 0, matot = 0, entot = 0, prtot = 0;
scanf("%d", &cid);
if (cid == 0) {
for (int i = 0; i < n; i++) {
if (removed[i] == 1) {
continue;
}
all++;
passnum = 0;
cntot = cntot + st[i].Ch;
matot = matot + st[i].Ma;
entot = entot + st[i].En;
prtot = prtot + st[i].Pr;
if (st[i].Ch >= 60) {
passnum++;
chnum++;
}
if (st[i].En >= 60) {
passnum++;
ennum++;
}
if (st[i].Ma >= 60) {
passnum++;
manum++;
}
if (st[i].Pr >= 60) {
passnum++;
prnum++;
}
if (passnum == 0) pass0++;
if (passnum >= 1) pass1++;
if (passnum >= 2) pass2++;
if (passnum >= 3) pass3++;
if (passnum >= 4) pass4++;
}
printf("Chinese\n"
"Average Score: %.2f\n"
"Number of passed students: %d\n"
"Number of failed students: %d\n\n"
"Mathematics\n"
"Average Score: %.2f\n"
"Number of passed students: %d\n"
"Number of failed students: %d\n\n"
"English\n"
"Average Score: %.2f\n"
"Number of passed students: %d\n"
"Number of failed students: %d\n\n"
"Programming\n"
"Average Score: %.2f\n"
"Number of passed students: %d\n"
"Number of failed students: %d\n\n"
"Overall:\n"
"Number of students who passed all subjects: %d\n"
"Number of students who passed 3 or more subjects: %d\n"
"Number of students who passed 2 or more subjects: %d\n"
"Number of students who passed 1 or more subjects: %d\n"
"Number of students who failed all subjects: %d\n\n",
cntot / (double)all + EPS, chnum, all - chnum,
matot / (double)all + EPS, manum, all - manum,
entot / (double)all + EPS, ennum, all - ennum,
prtot / (double)all + EPS, prnum, all - prnum,
pass4, pass3, pass2, pass1, pass0
);
} else {
for (int i = 0; i < n; i++) {
if (removed[i] == 1 || st[i].CID != cid) {
continue;
}
all++;
passnum = 0;
cntot = cntot + st[i].Ch;
matot = matot + st[i].Ma;
entot = entot + st[i].En;
prtot = prtot + st[i].Pr;
if (st[i].Ch >= 60) {
passnum++;
chnum++;
}
if (st[i].En >= 60) {
passnum++;
ennum++;
}
if (st[i].Ma >= 60) {
passnum++;
manum++;
}
if (st[i].Pr >= 60) {
passnum++;
prnum++;
}
if (passnum == 0) pass0++;
if (passnum >= 1) pass1++;
if (passnum >= 2) pass2++;
if (passnum >= 3) pass3++;
if (passnum >= 4) pass4++;
}
printf("Chinese\n"
"Average Score: %.2f\n"
"Number of passed students: %d\n"
"Number of failed students: %d\n\n"
"Mathematics\n"
"Average Score: %.2f\n"
"Number of passed students: %d\n"
"Number of failed students: %d\n\n"
"English\n"
"Average Score: %.2f\n"
"Number of passed students: %d\n"
"Number of failed students: %d\n\n"
"Programming\n"
"Average Score: %.2f\n"
"Number of passed students: %d\n"
"Number of failed students: %d\n\n"
"Overall:\n"
"Number of students who passed all subjects: %d\n"
"Number of students who passed 3 or more subjects: %d\n"
"Number of students who passed 2 or more subjects: %d\n"
"Number of students who passed 1 or more subjects: %d\n"
"Number of students who failed all subjects: %d\n\n",
cntot / (double)all + EPS, chnum, all - chnum,
matot / (double)all + EPS, manum, all - manum,
entot / (double)all + EPS, ennum, all - ennum,
prtot / (double)all + EPS, prnum, all - prnum,
pass4, pass3, pass2, pass1, pass0
);
}
}
int main() {
while (1) {
int choice;
// 注意输出格式!!!
printf("Welcome to Student Performance Management System (SPMS).\n");
printf("\n");
printf("1 - Add\n");
printf("2 - Remove\n");
printf("3 - Query\n");
printf("4 - Show ranking\n");
printf("5 - Show Statistics\n");
printf("0 - Exit\n");
printf("\n");
scanf("%d", &choice);
switch (choice) {
case 0:
return 0;
case 1:
add();
break;
case 2:
RQ(1);
break;
case 3:
RQ(0);
break;
case 4:
printf("Showing the ranklist hurts students' self-esteem. Don't do that.\n");
break;
case 5:
ST();
break;
}
}
return 0;
}
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