A1062 Talent and Virtue (25 分| 排序，附详细注释，逻辑分析)

写在前面

• 思路分析

• 结构体数组封装学生元信息

• 结构体数组vector v[4]封装4类考生

• cmp函数，较耗费时间

• 排序先按照总分排序，然后按照德分排序，最后按照才分排序

• 输出符合条件的结果～

• 题目简单，20分钟a题

测试用例

input:
14 60 80
10000001 64 90
10000002 90 60
10000011 85 80
10000003 85 80
10000004 80 85
10000005 82 77
10000006 83 76
10000007 90 78
10000008 75 79
10000009 59 90
10000010 88 45
10000012 80 100
10000013 90 99
10000014 66 60

output:
12
10000013 90 99
10000012 80 100
10000003 85 80
10000011 85 80
10000004 80 85
10000007 90 78
10000006 83 76
10000005 82 77
10000002 90 60
10000014 66 60
10000008 75 79
10000001 64 90

ac代码

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct node
{
    int num, de, cai;
};

int cmp(node a,node b)
{
    if((a.de + a.cai) != (b.de+b.cai))
        return (a.de + a.cai) > (b.de+b.cai);
    else if(a.de != b.de)
        return a.de>b.de;
    else
        return a.num<b.num;
}

int main()
{
    int n, low, high;
    scanf("%d %d %d", &n, &low, &high);
    vector<node> v[4];
    node tmp;
    int total = n;

    for(int i=0; i<n; i++)
    {
        scanf("%d %d %d", &tmp.num,&tmp.de,&tmp.cai);
        if(tmp.de < low || tmp.cai < low)
            total--;
        else if(tmp.de >= high && tmp.cai >= high)
            v[0].push_back(tmp);
        else if(tmp.de >= high && tmp.cai<high)
            v[1].push_back(tmp);
        else if(tmp.de < high && tmp.cai < high && tmp.de >= tmp.cai)
            v[2].push_back(tmp);
        else
            v[3].push_back(tmp);
    }

    printf("%d\n", total);
    for(int i=0; i<4; i++)
    {
        sort(v[i].begin(), v[i].end(), cmp);
        for(int j=0; j<v[i].size(); j++)
            printf("%d %d %d\n", v[i][j].num, v[i][j].de, v[i][j].cai);
    }

    return 0;
}