Codeforces Round #540 (Div. 3)--1118B - Tanya and Candies(easy TL!)
Tanya has nn candies numbered from 11 to nn. The ii-th candy has the weight aiai.
She plans to eat exactly n−1n−1 candies and give the remaining candy to her dad. Tanya eats candies in order of increasing their numbers, exactly one candy per day.
Your task is to find the number of such candies ii (let's call these candies good) that if dad gets the ii-th candy then the sum of weights of candies Tanya eats in even days will be equal to the sum of weights of candies Tanya eats in odd days. Note that at first, she will give the candy, after it she will eat the remaining candies one by one.
For example, n=4n=4 and weights are [1,4,3,3][1,4,3,3]. Consider all possible cases to give a candy to dad:
- Tanya gives the 11-st candy to dad (a1=1a1=1), the remaining candies are [4,3,3][4,3,3]. She will eat a2=4a2=4 in the first day, a3=3a3=3 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 4+3=74+3=7 and in even days she will eat 33. Since 7≠37≠3 this case shouldn't be counted to the answer (this candy isn't good).
- Tanya gives the 22-nd candy to dad (a2=4a2=4), the remaining candies are [1,3,3][1,3,3]. She will eat a1=1a1=1 in the first day, a3=3a3=3 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 33. Since 4≠34≠3 this case shouldn't be counted to the answer (this candy isn't good).
- Tanya gives the 33-rd candy to dad (a3=3a3=3), the remaining candies are [1,4,3][1,4,3]. She will eat a1=1a1=1 in the first day, a2=4a2=4 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 44. Since 4=44=4 this case should be counted to the answer (this candy is good).
- Tanya gives the 44-th candy to dad (a4=3a4=3), the remaining candies are [1,4,3][1,4,3]. She will eat a1=1a1=1 in the first day, a2=4a2=4 in the second day, a3=3a3=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 44. Since 4=44=4 this case should be counted to the answer (this candy is good).
In total there 22 cases which should counted (these candies are good), so the answer is 22.
The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of candies.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1041≤ai≤104), where aiai is the weight of the ii-th candy.
Print one integer — the number of such candies ii (good candies) that if dad gets the ii-th candy then the sum of weights of candies Tanya eats in even days will be equal to the sum of weights of candies Tanya eats in odd days.
7
5 5 4 5 5 5 6
2
8
4 8 8 7 8 4 4 5
2
9
2 3 4 2 2 3 2 2 4
3
#include<iostream>
using namespace std;
int num[];
int main(){
int n;
cin>>n;
int evenPre=,oddPre=,even=,odd=;
for(int i=;i<n;i++){
cin>>num[i];
if(i&)
odd+=num[i];
else
even+=num[i];
}
int ans=;
for(int i=;i<n;i++){
if(i&)
odd-=num[i];
else
even-=num[i];
if(oddPre+even==evenPre+odd)
ans++;
if(i&)
oddPre+=num[i];
else
evenPre+=num[i];
}
cout<<ans<<endl;
return ;
}
Ta
nn;njklya has nn candies numbered from 11 to nn. The ii-th candy has the weight aiai.
She plans to eat exactly n−1n−1 candies and give the remaining candy to her dad. Tanya eats candies in order of increasing their numbers, exactly one candy per day.
Your task is to find the number of such candies ii (let's call these candies good) that if dad gets the ii-th candy then the sum of weights of candies Tanya eats in even days will be equal to the sum of weights of candies Tanya eats in odd days. Note that at first, she will give the candy, after it she will eat the remaining candies one by one.
For example, n=4n=4 and weights are [1,4,3,3][1,4,3,3]. Consider all possible cases to give a candy to dad:
- Tanya gives the 11-st candy to dad (a1=1a1=1), the remaining candies are [4,3,3][4,3,3]. She will eat a2=4a2=4 in the first day, a3=3a3=3 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 4+3=74+3=7 and in even days she will eat 33. Since 7≠37≠3 this case shouldn't be counted to the answer (this candy isn't good).
- Tanya gives the 22-nd candy to dad (a2=4a2=4), the remaining candies are [1,3,3][1,3,3]. She will eat a1=1a1=1 in the first day, a3=3a3=3 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 33. Since 4≠34≠3 this case shouldn't be counted to the answer (this candy isn't good).
- Tanya gives the 33-rd candy to dad (a3=3a3=3), the remaining candies are [1,4,3][1,4,3]. She will eat a1=1a1=1 in the first day, a2=4a2=4 in the second day, a4=3a4=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 44. Since 4=44=4 this case should be counted to the answer (this candy is good).
- Tanya gives the 44-th candy to dad (a4=3a4=3), the remaining candies are [1,4,3][1,4,3]. She will eat a1=1a1=1 in the first day, a2=4a2=4 in the second day, a3=3a3=3 in the third day. So in odd days she will eat 1+3=41+3=4 and in even days she will eat 44. Since 4=44=4 this case should be counted to the answer (this candy is good).
In total there 22 cases which should counted (these candies are good), so the answer is 22.
The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of candies.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1041≤ai≤104), where aiai is the weight of the ii-th candy.
Print one integer — the number of such candies ii (good candies) that if dad gets the ii-th candy then the sum of weights of candies Tanya eats in even days will be equal to the sum of weights of candies Tanya eats in odd days.
7
5 5 4 5 5 5 6
2
8
4 8 8 7 8 4 4 5
2
9
2 3 4 2 2 3 2 2 4
3
Codeforces Round #540 (Div. 3)--1118B - Tanya and Candies(easy TL!)的更多相关文章
- Codeforces Round #540 (Div. 3) B. Tanya and Candies (后缀和)
题意:有\(n\)个数,你可以任意去除某个位置的元素然后得到一个新数组,使得新数组奇数位和偶数的元素相等,现在问你有多少种情况合法. 题解:先求个后缀和,然后遍历,记录奇数和偶数位置的前缀和,删去\( ...
- Codeforces Round #540 (Div. 3)--1118D1 - Coffee and Coursework (Easy version)
https://codeforces.com/contest/1118/problem/D1 能做完的天数最大不超过n,因为假如每天一杯咖啡,每杯咖啡容量大于1 首先对容量进行从大到小的排序, sor ...
- Codeforces Round #540 (Div. 3) D1. Coffee and Coursework (Easy version) 【贪心】
任意门:http://codeforces.com/contest/1118/problem/D1 D1. Coffee and Coursework (Easy version) time limi ...
- Codeforces Round #540 (Div. 3) 部分题解
Codeforces Round #540 (Div. 3) 题目链接:https://codeforces.com/contest/1118 题目太多啦,解释题意都花很多时间...还有事情要做,就选 ...
- Codeforces Round #540 (Div. 3) A,B,C,D2,E,F1
A. Water Buying 链接:http://codeforces.com/contest/1118/problem/A 实现代码: #include<bits/stdc++.h> ...
- Codeforces Round #540 (Div. 3) F1. Tree Cutting (Easy Version) 【DFS】
任意门:http://codeforces.com/contest/1118/problem/F1 F1. Tree Cutting (Easy Version) time limit per tes ...
- Codeforces Round #288 (Div. 2)D. Tanya and Password 欧拉通路
D. Tanya and Password Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/508 ...
- Codeforces Round #346 (Div. 2) C Tanya and Toys
C. Tanya and Toys 题目链接http://codeforces.com/contest/659/problem/C Description In Berland recently a ...
- Codeforces Round #346 (Div. 2) C. Tanya and Toys 贪心
C. Tanya and Toys 题目连接: http://www.codeforces.com/contest/659/problem/C Description In Berland recen ...
随机推荐
- c# mac地址 和http://xx.xx.xx/ 正则表达式匹配
Mac :^([0-9a-fA-F]{2})(([/\s:][0-9a-fA-F]{2}){5})$ C# 书写方式 一下是允许mac中间间隔符是“:”或者“-”两种输入方式 并且我把上边的正则表达 ...
- spring boot (一): Hello World
什么是spring boot Spring Boot是由Pivotal团队提供的全新框架,其设计目的是用来简化新Spring应用的初始搭建以及开发过程.该框架使用了特定的方式来进行配置,从而使开发人员 ...
- spring自动注解Autowired配置
1.spring注解:http://blog.csdn.net/xyh820/article/details/7303330/ 2.最简ssm配置:http://blog.csdn.net/qq_18 ...
- delphi sdk 函数个数知多少?
pascal用了这么久 那么您知道他有多少个函数,过程? 笔者统计了一下, delphi 7 21579个delphi xe2 41145个lazarus 1.12 70987个 ==== ...
- android xml解析中的null问题
当我们从服务器或者xml文件加载xml进行解析的时候,往往报告 nullpointer 错误.这是原始代码: String short_name = doc.getElementsByTagName( ...
- Maximum Swap LT670
Given a non-negative integer, you could swap two digits at most once to get the maximum valued numbe ...
- springboot server.address 配置问题
1. server.address 为对应机器ip地址时 ,如 18.10.x.x 此时访问该服务只能使用 ip 访问 . 2. 配置为 127.0.0.1 时 可以使用 localhost 和 ...
- 【UI测试】--多窗口&系统资源
- 嵌入式操作系统VxWorks中网络协议存储池原理及实现
嵌入式操作系统VxWorks中网络协议存储池原理及实现 周卫东 蔺妍 刘利强 (哈尔滨工程大学自动化学院,黑龙江 哈尔滨,150001) 摘 要 本文讨论了网络协议存储池的基本原理和在嵌入式操作系 ...
- OneZero第三周第五次站立会议(2016.4.8)
1. 时间: 15:10--15:25 共计15分钟. 2. 成员: X 夏一鸣 * 组长 (博客:http://www.cnblogs.com/xiaym896/), G 郭又铭 (博客:http ...