Educational Codeforces Round 31
2 seconds
256 megabytes
standard input
standard output
Recently Luba bought a very interesting book. She knows that it will take t seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of n next days. The number of seconds that Luba has to spend working during i-th day is ai. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed n.
Remember that there are 86400 seconds in a day.
The first line contains two integers n and t (1 ≤ n ≤ 100, 1 ≤ t ≤ 106) — the number of days and the time required to read the book.
The second line contains n integers ai (0 ≤ ai ≤ 86400) — the time Luba has to spend on her work during i-th day.
Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed n.
2 2
86400 86398
2
2 86400
0 86400
1
直接模拟下最好
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,k;
cin>>n>>k;
for(int i=;i<=n;i++)
{
int x;
cin>>x;
k-=-x;
if(k<=)
{
cout<<i<<endl;
return ;
}
}
return ;
}
Educational Codeforces Round 31的更多相关文章
- Educational Codeforces Round 31 B. Japanese Crosswords Strike Back【暴力】
B. Japanese Crosswords Strike Back time limit per test 1 second memory limit per test 256 megabytes ...
- Educational Codeforces Round 31 A. Book Reading【暴力】
A. Book Reading time limit per test 2 seconds memory limit per test 256 megabytes input standard inp ...
- 【Educational Codeforces Round 31 C】Bertown Subway
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 最后肯定会形成若干个环的. 把最大的两个环合在一起就好. 每个环贡献: 假设x=环的大小 ->x*x 注意int的溢出 [代码 ...
- 【Educational Codeforces Round 31 B】Japanese Crosswords Strike Back
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 所有数字的和加上n-1,如果为x则唯一,否则不唯一 [代码] #include <bits/stdc++.h> usin ...
- 【Educational Codeforces Round 31 A】Book Reading
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 水模拟 [代码] #include <bits/stdc++.h> using namespace std; const ...
- Educational Codeforces Round 32
http://codeforces.com/contest/888 A Local Extrema[水] [题意]:计算极值点个数 [分析]:除了第一个最后一个外,遇到极值点ans++,包括极大和极小 ...
- Educational Codeforces Round 17
Educational Codeforces Round 17 A. k-th divisor 水题,把所有因子找出来排序然后找第\(k\)大 view code //#pragma GCC opti ...
- [Educational Codeforces Round 16]E. Generate a String
[Educational Codeforces Round 16]E. Generate a String 试题描述 zscoder wants to generate an input file f ...
- [Educational Codeforces Round 16]D. Two Arithmetic Progressions
[Educational Codeforces Round 16]D. Two Arithmetic Progressions 试题描述 You are given two arithmetic pr ...
随机推荐
- jQuery.Deferred(jQuery1.5-2.1)源码剖析
jQuery.Deferred作为1.5的新特性出现在jQuery上,而jQuery.ajax函数也做了相应的调整.因此我们能如下的使用xhr请求调用,并实现事件处理函数晚绑定. var promis ...
- uvm_reg_item——寄存器模型(五)
uvm_reg_item 扩展自uvm_sequence_item,也就说寄存器模型定义了transaction item. adapter 的作用是把这uvm_reg_item转换成uvm_sequ ...
- 关于软件测试(5):初识Peer Review
一.背景:这周的软件测试课堂上我们在自行分组的情况下,对姚同学的汽车停车位定位管理系统进行了Peer Review,中文就是同行测试.这也是我第一次接触同行测试,那接下来我先介绍一下Peer Revi ...
- codevs 1131 统计单词数 2011年NOIP全国联赛普及组
时间限制: 1 s 空间限制: 128000 KB 题目等级 : 白银 Silver 题目描述 Description 一般的文本编辑器都有查找单词的功能,该功能可以快速定位特定单词在文章中的位 ...
- Nodejs + Jshint自动化静态代码检查
1. 目的 提交代码前能够自动化静态代码检查,提高代码质量 2. 准备 1. Nodejs安装: 官方地址:http://nodejs.org/ 安装说明:根据电脑配置下载对应的版本进行 ...
- Data truncation: Data too long for column 'id' at row 1
Caused by: java.sql.BatchUpdateException: Data truncation: Data too long for column 'titleimg' at ro ...
- 用dfs求联通块(UVa572)
一.题目 输入一个m行n列的字符矩阵,统计字符“@”组成多少个八连块.如果两个字符所在的格子相邻(横.竖.或者对角线方向),就说它们属于同一个八连块. 二.解题思路 和前面的二叉树遍历类似,图也有DF ...
- WPF知识点全攻略04- XAML页面布局
名称 说明 Canvas 使用固定坐标绝对定位元素 StackPanel 在水平或竖直方向放置元素 DockPanel 根据外部容器边界,自动调整元素 WrapPanel 在可换行的行中放置元素 Gr ...
- 配置SpringMVC返回JSON遇到的坑
坑一:官方网站下载地址不明朗,最后找了几个下载地址:http://wiki.fasterxml.com/JacksonDownload Jackson2.5下载地址:jackson2.5.0.jar ...
- POJ-2251-地下城
这题是一道简单的广搜题目,读入的时候,需要注意,如果是用scanf读入的话,就直接读取每行的字符串,不然的话,行尾的回车,也会被当成字符读入,这样的话,每次读取的数目就会小于我们想要的数目,因为每次把 ...