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117 lines
3.8 KiB
C
117 lines
3.8 KiB
C
/**
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* 1015. 德才论 (25)
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*
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* 宋代史学家司马光在《资治通鉴》中有一段著名的“德才论”:“是故才德全尽谓之圣人,
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* 才德兼亡谓之愚人,德胜才谓之君子,才胜德谓之小人。凡取人之术,苟不得圣人,君子
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* 而与之,与其得小人,不若得愚人。”
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*
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* 现给出一批考生的德才分数,请根据司马光的理论给出录取排名。
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*
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* 输入格式:
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*
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* 输入第1行给出3个正整数,分别为:N(<=10^5),即考生总数;L(>=60),为录取最低
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* 分数线,即德分和才分均不低于L的考生才有资格被考虑录取;H(<100),为优先录取线
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* ——德分和才分均不低于此线的被定义为“才德全尽”,此类考生按德才总分从高到低排序;
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* 才分不到但德分到线的一类考生属于“德胜才”,也按总分排序,但排在第一类考生之后;
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* 德才分均低于H,但是德分不低于才分的考生属于“才德兼亡”但尚有“德胜才”者,按总分
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* 排序,但排在第二类考生之后;其他达到最低线L的考生也按总分排序,但排在第三类考生之后。
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*
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* 随后N行,每行给出一位考生的信息,包括:准考证号、德分、才分,其中准考证号为8位
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* 整数,德才分为区间[0, 100]内的整数。数字间以空格分隔。
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*
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* 输出格式:
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*
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* 输出第1行首先给出达到最低分数线的考生人数M,随后M行,每行按照输入格式输出一位
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* 考生的信息,考生按输入中说明的规则从高到低排序。当某类考生中有多人总分相同时,按
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* 其德分降序排列;若德分也并列,则按准考证号的升序输出。
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*
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* 输入样例:
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* 14 60 80
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* 10000001 64 90
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* 10000002 90 60
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* 10000011 85 80
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* 10000003 85 80
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* 10000004 80 85
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* 10000005 82 77
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* 10000006 83 76
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* 10000007 90 78
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* 10000008 75 79
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* 10000009 59 90
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* 10000010 88 45
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* 10000012 80 100
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* 10000013 90 99
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* 10000014 66 60
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* 输出样例:
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* 12
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* 10000013 90 99
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* 10000012 80 100
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* 10000003 85 80
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* 10000011 85 80
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* 10000004 80 85
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* 10000007 90 78
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* 10000006 83 76
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* 10000005 82 77
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* 10000002 90 60
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* 10000014 66 60
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* 10000008 75 79
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* 10000001 64 90
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**/
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#include <stdio.h>
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#include <stdlib.h>
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typedef struct _Student{
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int ID;
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int D; /** de2 : virtue */
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int C; /** cai2: ability */
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int rank;
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int sum; /** sum = D + C */
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}sStudent, *Student;
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/* larger the number, higher the rank */
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int rank(Student s, int H, int L)
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{
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if(s->D < L || s->C < L) return 0; /* failed */
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else if(s->D >= H && s->C >= H) return 4; /* best */
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else if(s->D >= H) return 3; /* second */
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else if(s->D >= s->C) return 2; /* third */
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else return 1; /* fourth */
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}
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int comp(const void *a, const void *b)
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{
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Student s1 = *(Student*)a;
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Student s2 = *(Student*)b;
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if(s1->rank != s2->rank) return s1->rank - s2->rank;
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else if(s1->sum != s2->sum) return s1->sum - s2->sum;
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else if(s1->D != s2->D) return s1->D - s2->D;
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else if(s1->ID != s2->ID) return s2->ID - s1->ID;
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else return 0;
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}
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int main()
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{
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int N, L, H, M = 0;
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Student students[100000] = {0};
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sStudent buffer[100000];
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scanf("%d %d %d", &N, &L, &H);
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for(int i = 0; i < N; i++)
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{
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Student s = buffer + i;
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scanf("%d %d %d", &s->ID, &s->D, &s->C);
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s->sum = s->D + s->C;
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if((s->rank = rank(s, H, L)) != 0) /* record if passed */
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students[M++] = s;
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}
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qsort(students, M, sizeof(Student), comp);
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printf("%d\n", M);
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for(int i = M - 1; i >= 0; i--)
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printf("%d %d %d\n", students[i]->ID, students[i]->D, students[i]->C);
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return 0;
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}
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