发布于北京 牛牛的绩点
https://www.nowcoder.com/practice/de8bc3e736164786b07a4dd64ed172f7
from sys import stdin
read1 = stdin.read().split()
s = 0
n = 0
dict1 = {"A":4,"B":3,"C":2,"D":1,"F":0}
for i in range(len(read1)//2):
s += dict1[read1[2*i]]*int(read1[2*i+1])
n += int(read1[2*i+1])
print(f"{s/n:.2f}")
相关推荐
球球与墩墩：这不是前端常考的对象扁平化吗，面试官像是前端出来的
const flattern = (obj) => {
const res = {};
const dfs = (curr, path) => {
if(typeof curr === 'object' && curr !== null) {
const isArray = Array.isArray(curr);
for(let key in curr) {
const newPath = path ? isArray ? `${path}[${key}]` : `${path}.${key}` : key;
dfs(curr[key], newPath);
}
} else {
res[path] = curr
}
}
dfs(obj);
return res;
}
查看3道真题和解析 