08-1. Talent and Virtue

About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage（圣人）"; being less excellent but with one's virtue outweighs talent can be called a "nobleman（君子）"; being good in neither is a "fool man（愚人）"; yet a fool man is better than a "small man（小人）" who prefers talent than virtue.

Now given the grades of talent and virtue of a group of people, you are supposed to rank them according to Sima Guang's theory.

Input Specification:

Each input file contains one test case. Each case first gives 3 positive integers in a line: N (<=10⁵), the total number of people to be ranked; L (>=60), the lower bound of the qualified grades -- that is, only the ones whose grades of talent and virtue are both not below this line will be ranked; and H (<100), the higher line of qualification -- that is, those with both grades not below this line are considered as the "sages", and will be ranked in non-increasing order according to their total grades. Those with talent grades below H but virtue grades not are cosidered as the "noblemen", and are also ranked in non-increasing order according to their total grades, but they are listed after the "sages". Those with both grades below H, but with virtue not lower than talent are considered as the "fool men". They are ranked in the same way but after the "noblemen". The rest of people whose grades both pass the L line are ranked after the "fool men".

Then N lines follow, each gives the information of a person in the format:

ID_Number Virtue_Grade Talent_Grade

where ID_Number is an 8-digit number, and both grades are integers in [0, 100]. All the numbers are separated by a space.

Output Specification:

The first line of output must give M (<=N), the total number of people that are actually ranked. Then M lines follow, each gives the information of a person in the same format as the input, according to the ranking rules. If there is a tie of the total grade, they must be ranked with respect to their virtue grades in non-increasing order. If there is still a tie, then output in increasing order of their ID's.

Sample Input:

14 60 8010000001 64 9010000002 90 6010000011 85 8010000003 85 8010000004 80 8510000005 82 7710000006 83 7610000007 90 7810000008 75 7910000009 59 9010000010 88 4510000012 80 10010000013 90 9910000014 66 60

Sample Output:

1210000013 90 9910000012 80 10010000003 85 8010000011 85 8010000004 80 8510000007 90 7810000006 83 7610000005 82 7710000002 90 6010000014 66 6010000008 75 7910000001 64 90

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提示：英文太多，表示看不懂，所以请看中文版，http://www.patest.cn/contests/pat-b-practise/1015

分析：建立四个容器vector存放4个类别的容器，分别是：圣人、君子、愚人、小人。 然后直接用sort函数进行排序，自己写一个比较函数，下面代码有。

特比需要说明一下的是，这道题一开始我用c++做完是拿去测试，测试点3，4就是会测试超时，改了很久还是不行，最后通过相关的资料才明白是c++的输入输出的问题，

使用cout和cin会使3，4超时，但是改成c语言的printf和scanf就就解决了，所以在这里需要注意一下

源码：

#include<iostream>#include<algorithm>#include<vector>#include <cstdio>using namespace std;struct student {int ID_Number;int Virtue_Grade;int Talent_Grade;student(int id, int vg, int tg) :ID_Number(id), Virtue_Grade(vg), Talent_Grade(tg) {}void output(){printf("%d %d %d\n", ID_Number, Virtue_Grade, Talent_Grade);}};//比较函数bool compare(const student &m1, const student &m2){if ((m1.Talent_Grade + m1.Virtue_Grade) != (m2.Talent_Grade + m2.Virtue_Grade))return (m1.Talent_Grade + m1.Virtue_Grade) > (m2.Talent_Grade + m2.Virtue_Grade);else{if (m1.Virtue_Grade != m2.Virtue_Grade)return m1.Virtue_Grade > m2.Virtue_Grade;elsereturn m1.ID_Number < m2.ID_Number;}}int main(){int N, L, H;scanf("%d %d %d", &N, &L, &H);vector<student> s1, s2, s3, s4;int id,vg,tg;int cnt = 0;//读入信息并删掉成绩不符合要求的考生for (int i = 0; i < N; i++){scanf("%d %d %d", &id, &vg, &tg);if (tg < L || vg < L)//不合格者continue;//才德全尽 的考生放到s1if (tg >= H && vg >= H){s1.push_back(student(id,vg,tg));}//德胜才  的考生放到s2else if (vg >= H && tg <= H){s2.push_back(student(id, vg, tg));}//才德兼亡 但尚有 德胜才 的考生放到s3else if (vg < H && tg< H && vg >= tg){s3.push_back(student(id, vg, tg));}//剩下的合格考生放道s4elses4.push_back(student(id, vg, tg));cnt++;}cout << cnt << endl;;//稳定排序sort(s1.begin(), s1.end(), compare);sort(s2.begin(), s2.end(), compare);sort(s3.begin(), s3.end(), compare);sort(s4.begin(), s4.end(), compare);//输出for (int i = 0; i < s1.size(); i++){s1[i].output();}for (int i = 0; i < s2.size(); i++){s2[i].output();}for (int i = 0; i < s3.size(); i++){s3[i].output();}for (int i = 0; i < s4.size(); i++){s4[i].output();}return 0;}