读入三科成绩算总分，按总分、语文、学号排序后输出前 5 名。
OJ: roj
题目 ID: 3552
难度:入门
标签:入门条件判断python
创建: 2026-10-02 07:06
更新: 2026-10-06 13:43
题目描述
思路
读入时计算总分并记录语文成绩和学号，用结构体数组按规则排序后输出前
参考代码
Python 版：
python
#!/usr/bin/env python3
# Author by Rainboy blog: https://rainboylv.com github: https://github.com/rainboylvx
# rbook: -> https://rbook.roj.ac.cn https://rbook2.roj.ac.cn
# rainboy的学习导航网站: https://idx.roj.ac.cn
# create_at: 2026-10-02 07:06
# update_at: 2026-10-02 07:45
import sys
def money(avg: float, ev: float, cadre: str, west: str, papers: int) -> int:
"""五项奖学金全部命中的金额之和：bool 乘金额，命中加钱、未命中加 0。"""
return (
8000 * (avg > 80 and papers >= 1) # 均分 >80 且有论文
+ 4000 * (avg > 85 and ev > 80) # 均分 >85 且评议 >80
+ 2000 * (avg > 90) # 均分 >90
+ 1000 * (avg > 85 and west == "Y") # 均分 >85 的西部学生
+ 850 * (ev > 80 and cadre == "Y") # 评议 >80 的学生干部
)
def solve() -> None:
lines = sys.stdin.read().splitlines()
n = int(lines[0]) # 学生人数
top_name, top_money, total = "", -1, 0
for line in lines[1 : n + 1]:
name, avg, ev, cadre, west, papers = line.split()
amount = money(float(avg), float(ev), cadre, west, int(papers))
total += amount
if amount > top_money: # 严格大于：并列时保留最先出现的学生
top_name, top_money = name, amount
print(f"{top_name}\n{top_money}\n{total}")
if __name__ == "__main__":
solve()C++ 版（同一算法）：
cpp
/**
* Author by Rainboy blog: https://rainboylv.com github: https://github.com/rainboylvx
* rbook: -> https://rbook.roj.ac.cn https://rbook2.roj.ac.cn
* rainboy的学习导航网站: https://idx.roj.ac.cn
* create_at: 2026-10-06 13:43
* update_at: 2026-10-06 13:45
*/
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 305; // 学生人数上限
struct Student {
int id; // 学号
int total; // 三科总分
int chinese; // 语文成绩
} a[MAXN];
int n;
// 排序规则：总分降序，语文降序，学号升序
bool cmp(const Student &x, const Student &y) {
if (x.total != y.total) return x.total > y.total;
if (x.chinese != y.chinese) return x.chinese > y.chinese;
return x.id < y.id;
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
int c, m, e;
scanf("%d%d%d", &c, &m, &e);
a[i].id = i;
a[i].chinese = c;
a[i].total = c + m + e;
}
sort(a + 1, a + n + 1, cmp);
int k = n < 5 ? n : 5; // 输出前 5 名，不足 5 名则全部输出
for (int i = 1; i <= k; i++) {
printf("%d %d\n", a[i].id, a[i].total);
}
return 0;
}