UVA11059 - Maximum Product
1、题目名称
Maximum Product
2、题目地址
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2000
3、题目内容
Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of the maximum positive product involving consecutive terms of S. If you cannot find a positive sequence, you should consider 0 as the value of the maximum product.
Input
Each test case starts with 1 ≤ N ≤ 18, the number of elements in a sequence. Each element Si is an integer such that −10 ≤ Si ≤ 10. Next line will have N integers, representing the value of each element in the sequence. There is a blank line after each test case. The input is terminated by end of file (EOF).
Output
For each test case you must print the message: ‘Case #M: The maximum product is P.’, where M is the number of the test case, starting from 1, and P is the value of the maximum product. After each test case you must print a blank line.
Sample Input
3
2 4 -3
5
2 5 -1 2 -1
Sample Output
Case #1: The maximum product is 8.
Case #2: The maximum product is 20.
大致意思就是,给出一个序列,问这个序列中最大连续累乘的子序列中,最大的值为多少,如果都为负数,则输出0.
感受: 一定要记得用long long , 还有格式问题,否则可能pe
#include"iostream"
using namespace std;
int a[];
int main(){
int n,out;
out=;
while(cin>>n){
for(int i=;i<n;i++)
cin>>a[i];
long long max=;
for(int i=;i<n;i++)
for(int j=i;j<n;j++){
long long z=;
for(int q=i;q<=j;q++)
z=z*a[q];
if(max<z) max=z;
}
cout<<"Case #"<<++out<<": The maximum product is "<<max<<"."<<endl<<endl; }
}
UVA11059 - Maximum Product的更多相关文章
- uva 11059 maximum product(水题)——yhx
aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAB1QAAAMcCAIAAABo0QCJAAAgAElEQVR4nOydW7msuhKF2wIasIAHJK
- [LeetCode] Maximum Product Subarray 求最大子数组乘积
Find the contiguous subarray within an array (containing at least one number) which has the largest ...
- 求连续最大子序列积 - leetcode. 152 Maximum Product Subarray
题目链接:Maximum Product Subarray solutions同步在github 题目很简单,给一个数组,求一个连续的子数组,使得数组元素之积最大.这是求连续最大子序列和的加强版,我们 ...
- Uva 11059 Maximum Product
注意long long long long longlong !!!!!! 还有 printf的时候 明明longlong型的答案 用了%d WA了也看不出,这个细节要注意!!! #incl ...
- LeetCode: Maximum Product Subarray && Maximum Subarray &子序列相关
Maximum Product Subarray Title: Find the contiguous subarray within an array (containing at least on ...
- 最大乘积(Maximum Product,UVA 11059)
Problem D - Maximum Product Time Limit: 1 second Given a sequence of integers S = {S1, S2, ..., Sn}, ...
- 暴力求解——最大乘积 Maximum Product,UVa 11059
最大乘积 Maximum Product 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=84562#problem/B 解题思路 ...
- LeetCode Maximum Product Subarray(枚举)
LeetCode Maximum Product Subarray Description Given a sequence of integers S = {S1, S2, . . . , Sn}, ...
- LeetCode 628. Maximum Product of Three Numbers (最大三数乘积)
Given an integer array, find three numbers whose product is maximum and output the maximum product. ...
随机推荐
- python pymysql安装
==================pymysql=================== 由于 MySQLdb 模块还不支持 Python3.x,所以 Python3.x 如果想连接MySQL需要安装 ...
- kafka的并行度与JStorm性能优化
kafka的并行度与JStorm性能优化 > Consumers Messaging traditionally has two models: queuing and publish-subs ...
- Mysql 行数据转换为列数据
现有如下表: 需要统计手机所有售卖价格,显示为如下表: 需要使用group_concat对price进行处理,并且去重重复价格 sql如下: select type,group_concat(DIST ...
- OKR与KPI管理的区别与联系
OKR是一种新兴的管理体系,最近几年被引进中国.由于在IT.互联网.金融.游戏等知识密集型企业中有着显著的效果,得到中国企业的认可. OKR是英文Objectives & Key Result ...
- NumPy入门基础【2】
通用函数ufunc 一元ufunc举例: 1.abs.fabs:计算绝对值,fabs更快 2.sqrt:计算各元素的平方根,相当于arr0.5 3.square:计算各元素的平方根,相当远arr2 4 ...
- Protobuf 语法 - 史上最简教程
Protobuf 语法简明教程 疯狂创客圈 死磕Netty 亿级流量架构系列之12 [博客园 总入口 ] 在protobuf中,协议是由一系列的消息组成的.因此最重要的就是定义通信时使用到的消息格式. ...
- 【python】-- MySQL简介、安装、操作
MySQL简介.安装.操作 数据库(Database)是按照数据结构来组织.存储和管理数据的仓库,每个数据库都有一个或多个不同的API用于创建,访问,管理,搜索和复制所保存的数据.我们也可以将数据存储 ...
- react-native 使用 antd-mobile-rn UI进行开发app
1.创建 react-native 项目 react-native init app03 2.安装组件 npm install antd-mobile-rn --save 3.配置按需加载 npm i ...
- JavaScript 中 onload 事件绑定多个方法的优化建议
页面加载完毕时会触发 onload 事件.基于内容(HTML)要与行为(JavaScript)分离的编码思想,我们需要将一些对页面的初始化操作写在方法内,并通过window.onload = func ...
- ABAP操作EXCEL (号称超级版)
[转自http://www.cnblogs.com/VerySky/articles/2170014.html] *------------------------------------------ ...