题目传送门

import math as m
ans = 0
n = int(input())
for i in range(1,n+1):
    a = int(input())
    sum = m.ceil(a / 70) * 0.1
    ans += sum
print("%.1f"%(ans))

#include <bits/stdc++.h> using namespace std; int main() { int a; double b=0; cin >> a; for(int i=1;i<=a;i++) { int xx; cin >> xx; b+=(xx+70-1)/70*0.1;

}
cout << fixed << setprecision(1) << b << endl;
return 0;

}

https://oj.jifu.nz/p/P172

import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans))

import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans))

import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans))

#include <bits/stdc++.h> using namespace std; int main() { cout#P172. 练28.3 短信计费 ID: 197 远端评测题 1000ms 256MiB 尝试: 31 已通过: 10 难度: 6 上传者:

jifu 标签> 说明 用手机发短信，一条短信资费为 0.1 0.1元，但限定一条短信的内容在 70 70 个字以内（包括 70 70 个字）。如果一次所发送的短信超过了 70 70 个字，则会按照每 70 70 个字一条短信的限制把它分割成多条短信发送。假设已经知道某人当月所发送的短信的字数，试统计一下他当月短信的总资费。

输入格式 第一行是整数 ? n( 1 ≤ ? ≤ 100 1≤n≤100)，表示当月发送短信的总次数，接着 ? n 行每行一个整数(不超过 1000 1000)，表示每次短信的字数。

输出格式 输出一行，当月短信总资费，单位为元，精确到小数点后 1 1 位。

样例 输入数据 1 10 39 49 42 61 44 147 42 72 35 46 输出数据 1 1.3 进入在线编程模式 (Alt+E) 递交 讨论 (0) 题解 (1) 文件 统计 复制 状态 评测队列 服务状态 开发 开源 支持 帮助 联系我们 关于 关于 隐私 服务条款 版权申诉 Language 兼容模式 主题
return 0; }

https://oj.jifu.nz/user/213

3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342 7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235 4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837 2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035 2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904 2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787 6611195909 2164201989

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ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans)) import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans)) import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans)) import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans)) import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans)) import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans))import math as m ans = 0 n = int(input()) for i in range(1,n+1): a = int(input()) sum = m.ceil(a / 70) * 0.1 ans += sum print("%.1f"%(ans)) import math as m ans = 0 n = 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