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题目链接

代码:

#include<cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
//N<=40000,M<=100,k<=5
int now;
struct Stu {
    int ge, gi, g;
    int sch[6];
    int rank, id;
};
struct Sch {
    int quota;
    int admit[40005];
    int na;
    int lastone;
};
Sch school[105];
Stu stu[40005];
bool cmp(Stu a, Stu b) {
    if (a.g != b.g)return a.g > b.g;
    else if (a.ge != b.ge)return a.ge > b.ge;
    else return a.id < b.id;
}
bool cmp2(int a, int b) {
    return a < b;
}
int main() {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 0; i < m; i++) {
        scanf("%d", &school[i].quota);
        school[i].na = 0;
        school[i].lastone = -1;
    }
    for (int i = 0; i < n; i++) {
        stu[i].id = i;
        scanf("%d %d", &stu[i].ge, &stu[i].gi);
        for (int j = 0; j < k; j++) {
            scanf("%d", &stu[i].sch[j]);
        }
        stu[i].g = stu[i].ge + stu[i].gi;
    }
    sort(stu, stu + n, cmp);
    //排名
    for (int i = 0; i < n; i++) {
        if (i == 0)stu[i].rank = 1;
        else if (stu[i].g == stu[i - 1].g&&stu[i].ge == stu[i - 1].ge)
            stu[i].rank = stu[i - 1].rank;
        else stu[i].rank = i + 1;
    }
    //录取工作
    for (int i = 0; i < n; i++) {
        int s,lastone;
        for (int j = 0; j < k; j++) {
            s = stu[i].sch[j];
            lastone = school[s].lastone;
            if (school[s].quota != 0) {
                school[s].quota--;
                school[s].admit[school[s].na++] = stu[i].id;
                school[s].lastone = i;
                break;
            }
            else if (lastone != -1&& stu[i].rank == stu[lastone].rank) {
                school[s].admit[school[s].na++] = stu[i].id;
                school[s].lastone = i;
                break;
            }
        }
    }
    //输出
    for (int i = 0; i < m; i++) {
        if (school[i].na){
            int na = school[i].na;
            sort(school[i].admit, school[i].admit + na, cmp2);
            for (int j = 0; j < na; j++) {
                if (j == 0) {
                    printf("%d", school[i].admit[j]);
                }
                else {
                    printf(" %d", school[i].admit[j]);
                }
            }
        }
        printf("\n");
    }
    return 0;
}


Agent 智能体学习指南:从入门到求职的完整路线 AI Agent 不是更长的 prompt,而是一类系统设计——模型在受控环境里观察状态、选择动作、调用工具、接收反馈。本文系统梳理 Agent 的 5 个核心概念、7 个关键模块(Goal/State/Context/Tools/Loop/Guardrails/Eval)、三条学习路线(工程落地/研究算法/求职作品)、框架选型策略(LangGraph/OpenAI Agents SDK/Pydantic AI 等场景推荐),以及可执行的 7 天最小项目计划(从概念到简历 bullet)。 阅读详情

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基于微软的DirectShow.dll的二次开发1.自动识别连接的摄像头跟麦克风且可以对其进行选择(默认选中第一个)2.可录制wmv和avi两种格式的视频并且都带有音频,avi可以录制的分辨率和帧数进行选择,wmv默认进行了淸晰度优化如不喜欢可去除代码中有相关提示3.可在预览点开之后进行拍照,也可边录制视频边拍照4.添加了视频和音频的压缩功能(酌情使用)这次二次开发发现C#对DirectShow的相关信息太少,这次做完也不敢藏私共享出来,还望大家指正

PAT A1080 Graduate Admission (30 分) 排序

题目大意: 每名考生的成绩由 GE、GI和GF组成,GF = (GE + GI) / 2。排名时,先按 GF降序排列,如果 GF 相同,则按 GE 降序排列,如果 GF 和 GE 都相同,那么这些考生的排名相同。然后,按照排名顺序依次进行录取。每个考生有 K个志愿,按照顺序依次检查每个志愿学校,如果学校当前的招生数小于它的招生限额,就录取该名考生,否则检查下一个志愿。但是,如果当前学校录取...

HackingGate的博客 263

pat 甲级 排序 A1080 Graduate Admission (30分)

1080 Graduate Admission (30分) It is said that in 2011, there are about 100 graduate schools ready to proceed over 40,000 applications in Zhejiang Province. It would help a lot if you could write a program to automate the admission procedure. Each applica

weixin_42358356的博客 186

PAT-A1080. Graduate Admission (30)(排序)

结构体要写好。 #include #include #include #include #include using namespace std; #define MAX 40000+10 struct Student { int Ge, Gi, sum; int r; int choice[6]; int d; int stuid; } st

WJ的专栏 677

【算法笔记题解】PAT A.1080 Graduate Admission (30 分)

全文目录 前言 昨天刷的那个被牛客网坑了一把,今天PAT恢复了,喜大普奔啊。今天这道题比较简单。 题目描述 1080 Graduate Admission (30 分)https://pintia.cn/problem-sets/994805342720868352/problems/994805387268571136 It is said that in 2011, there are about 100 graduate schools ready to proceed over 40,00

XingleiGao的博客 395

PAT A1080 Graduate Admission (30)

It is said that in 2013, there were about 100 graduate schools ready to proceed over 40,000 applications in Zhejiang Province. It would help a lot if you could write a program to automate the admissio...

fried123123的博客 221

pat-A1080. Graduate Admission (30)

今天开始写博客 水题一个,不解释题目。注意输出时数字从小到大。然后就直接过了。代码如下: #include #include #include #include using namespace std; int num[105]; struct node{     int ge=0;     int gi=0;     int id;     vector add

dream_never_giveup的博客 326

PAT A1080 graduate admission (30)

#include #include #include using namespace std; struct stut{ int ge,gi,totl,uid; int apply[5]; int schl; int rank; }stu[40010]; struct sch{ int er; int stuid[40010]; int num; }school[102

momottyy的专栏 308

PAT A 1080. Graduate Admission (30)

题目

xyzchenzd的专栏 769

【PAT甲级A1080】Graduate Admission (30分)(c++)

1080 Graduate Admission (30分) 作者:CHEN, Yue 单位:浙江大学 代码长度限制:16 KB 时间限制:250 ms 内存限制:64 MB It is said that in 2011, there are about 100 graduate schools ready to proceed over 40,000 applications in Zhejiang Province. It would help a lot if you could write a p

小庄同学的博客 173
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