原题很长，就此略过了
/problem/UVA-12412#author=0
这道题看似是一道简单的家庭作业题，其实非常容易出错，且调试时间很长。主要的难点在于：
1. 增加学生成绩记录时调整好整个学生列表的排名Rank值，不用排序
2. 删除学生成绩记录后需要调整学生排名
3. 如何快速高效生成统计数据
#include<stdio.h>#include<string.h>#define EPS 1e-6const int max1 = 512;const int max2 = 85;char sid[max1][max2], name[max1][max2];int cid[max1], score[max1][5], rank[max1], removed[max1];int n = 0;void print_students() {printf("Current student num: %d\n", n);for(int i = 0; i < n; i++) {printf("Student %d: sid = %s cid = %d name = %s rank = %d removed = %d score[0] = %d score[1] = %d score[2] = %d score[3] = %d\n",i, sid[i], cid[i], name[i], rank[i], removed[i], score[i][0], score[i][1], score[i][2], score[i][3]);}printf("\n");}void print_menu() {printf("Welcome to Student Performance Management System (SPMS).\n\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\n");}void add() {char s[max2];bool dup;for(;;) {printf("Please enter the SID, CID, name and four scores. Enter 0 to finish.\n");dup = false;//print_students();scanf("%s", s);if(strcmp(s, "0") == 0) break;strcpy(sid[n], s);scanf("%d%s%d%d%d%d", &cid[n], name[n], &score[n][0], &score[n][1], &score[n][2], &score[n][3]);for(int i = 0; i < n; i++) {if(!removed[i] && strcmp(sid[i], s) == 0) {printf("Duplicated SID.\n");dup = true;break;}}if(dup) continue;score[n][4] = score[n][0] + score[n][1] + score[n][2] + score[n][3];int cnt = 0; // cnt记录成绩比score[n][4]更高的学生数量for(int i = 0; i < n; i++) {if(!removed[i]) {if(score[i][4] < score[n][4]) rank[i]++;if(score[i][4] > score[n][4]) cnt++;}}rank[n] = cnt + 1;n++;//print_students();}}void DQ(int isq){char s[max2];for(;;) {printf("Please enter SID or name. Enter 0 to finish.\n");scanf("%s", s);if(strcmp(s, "0") == 0) return;int r = 0; // record the num of removed studentsfor(int i = 0; i < n; i++) {if(!removed[i]) {if(strcmp(sid[i], s) == 0 || strcmp(name[i], s) == 0) {if(isq) // For queryprintf("%d %s %d %s %d %d %d %d %d %.2f\n", rank[i], sid[i], cid[i], name[i], score[i][0],score[i][1], score[i][2], score[i][3], score[i][4], score[i][4]/4.0+EPS);else {removed[i] = 1;r++;for(int j = 0; j < n; j++)if(!removed[j] && rank[j] > rank[i]) rank[j]--;}}}}if(!isq) printf("%2d student(s) removed.\n", r);}}void stat() {printf("Please enter class ID, 0 for the whole statistics.\n");int c;scanf("%d", &c);int cnt = 0, pass4=0, pass3=0, pass2=0, pass1=0, pass0=0;int sum[4], pass[4];memset(sum, 0, sizeof(sum));memset(pass, 0, sizeof(pass));for(int i = 0; i < n; i++) {if(!removed[i]) {if(c == 0 || cid[i] == c) {cnt++;int pass_cnt = 0;for(int k = 0; k < 4; k++) {sum[k] += score[i][k];if(score[i][k] >= 60) { pass[k]++; pass_cnt++; }}if(pass_cnt == 4) pass4++;if(pass_cnt == 3) pass3++;if(pass_cnt == 2) pass2++;if(pass_cnt == 1) pass1++;if(pass_cnt == 0) pass0++;} }}for(int k = 0; k < 4; k++) {// k 是课程编号if(k == 0) printf("%s\n", "Chinese");if(k == 1) printf("%s\n", "Mathematics");if(k == 2) printf("%s\n", "English");if(k == 3) printf("%s\n", "Programming");if(cnt > 0) printf("Average Score: %.2f\n", (float)sum[k]/cnt);else printf("Average Score: %.2f\n", (float)0.00);printf("Number of passed students: %d\n", pass[k]);printf("Number of failed students: %d\n\n", cnt - pass[k]);}printf("Overall:\n");printf("Number of students who passed all subjects: %d\n", pass4);printf("Number of students who passed 3 or more subjects: %d\n", pass4 + pass3);printf("Number of students who passed 2 or more subjects: %d\n", pass4 + pass3 + pass2);printf("Number of students who passed 1 or more subjects: %d\n", pass4 + pass3 + pass2 + pass1);printf("Number of students who failed all subjects: %d\n\n", pass0);}int main() {memset(removed, 0, sizeof(removed));for(;;) {int choice;print_menu();scanf("%d", &choice);if(choice == 0) break;if(choice == 1) add();if(choice == 2) DQ(0); // Remove student recordif(choice == 3) DQ(1); // Query student recordif(choice == 4) printf("Showing the ranklist hurts students' self-esteem. Don't do that.\n");if(choice == 5) stat();}return 0;}
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