Codeforces Round #274 (Div. 2) --A Expression
主题链接:Expression
1 second
256 megabytes
standard input
standard output
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers a, b, c on
the blackboard. The task was to insert signs of operations '+' and '*',
and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7
- 1*(2+3)=5
- 1*2*3=6
- (1+2)*3=9
Note that you can insert operation signs only between a and b,
and between b and c, that is, you cannot swap integers.
For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given a, b and c print
the maximum value that you can get.
The input contains three integers a, b and c,
each on a single line (1 ≤ a, b, c ≤ 10).
Print the maximum value of the expression that you can obtain.
1
2
3
9
2
10
3
60
大致题意:a, b, c三个数。在三个数中,插入“+” 和“*”运算符的随意两个组合,求能组成的表达式的值得最大值。(能够用括号)
解题思路:没啥说的。直接暴力,总共就6种组合。
AC代码:
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff int x[9]; int main()
{
// #ifdef sxk
// freopen("in.txt","r",stdin);
// #endif
int a,b,c;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
x[0] = a + b + c;
x[1] = a + (b * c);
x[2] = a * (b + c);
x[3] = (a + b) * c;
x[4] = (a * b) + c;
x[5] = a * b * c;
sort(x, x+6);
printf("%d\n",x[5]);
}
return 0;
}
版权声明:本文sxk原创文章。转载本文,请添加链接^_^
Codeforces Round #274 (Div. 2) --A Expression的更多相关文章
- codeforces水题100道 第八题 Codeforces Round #274 (Div. 2) A. Expression (math)
题目链接:http://www.codeforces.com/problemset/problem/479/A题意:给你三个数a,b,c,使用+,*,()使得表达式的值最大.C++代码: #inclu ...
- Codeforces Round #274 (Div. 2) 解题报告
题目地址:http://codeforces.com/contest/479 这次自己又仅仅能做出4道题来. A题:Expression 水题. 枚举六种情况求最大值就可以. 代码例如以下: #inc ...
- Codeforces Round #274 (Div. 2)
A http://codeforces.com/contest/479/problem/A 枚举情况 #include<cstdio> #include<algorithm> ...
- Codeforces Round #274 (Div. 1) C. Riding in a Lift 前缀和优化dp
C. Riding in a Lift Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/pr ...
- Codeforces Round #274 (Div. 1) B. Long Jumps 数学
B. Long Jumps Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/ ...
- Codeforces Round #274 (Div. 1) A. Exams 贪心
A. Exams Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/A Des ...
- Codeforces Round #274 (Div. 2)-C. Exams
http://codeforces.com/contest/479/problem/C C. Exams time limit per test 1 second memory limit per t ...
- Codeforces Round #274 Div.1 C Riding in a Lift --DP
题意:给定n个楼层,初始在a层,b层不可停留,每次选一个楼层x,当|x-now| < |x-b| 且 x != now 时可达(now表示当前位置),此时记录下x到序列中,走k步,最后问有多少种 ...
- Codeforces Round #274 (Div. 2) E. Riding in a Lift(DP)
Imagine that you are in a building that has exactly n floors. You can move between the floors in a l ...
随机推荐
- 【c语言】模拟库函数strstr
// 模拟库函数strstr #include <stdio.h> #include <assert.h> const char* my_strstr(const char * ...
- HTML5 CSS3专题 纯CSS打造相冊效果
转载请标明出处:http://blog.csdn.net/lmj623565791/article/details/30993277 今天偶然发现电脑里面还有这种一个样例.感觉效果还不错,不记得啥时候 ...
- php-GD库的函数(二)
<?php //imagecopy — 拷贝图像的一部分粘贴到某图像上 /*bool imagecopy ( resource $dst_im , resource $src_im , int ...
- Beginning Python From Novice to Professional (4) - 演示样本格式字符串
$ gedit price.py #!/usr/bin/env python width = input('Please enter width: ') price_width = 10 item_w ...
- CentOS 6.5安全加固及性能优化
(文章来自:http://www.cnblogs.com/seasonzone/p/3526296.html) 我们可以通过调整系统参数来提高系统内存.CPU.内核资源的占用,通过禁用不必要的服务.端 ...
- 【spring源代码分析】--Bean的解析与注冊
接着上一节继续分析,DefaultBeanDefinitionDocumentReader的parseBeanDefinitions方法: protected void parseBeanDefini ...
- 从零開始学习OpenCL开发(一)架构
多谢大家关注 转载本文请注明:http://blog.csdn.net/leonwei/article/details/8880012 本文将作为我<从零開始做OpenCL开发>系列文章的 ...
- AndroidUI组件之ListView小技巧
android:fadingEdge="none"//出去黑影 android:listSelector="@android:color/transparent&quo ...
- Effective C++:规定20: 宁pass-by-reference-to-const更换pass-by-value
(一) 假设传递参数当函数被调用pass-by-value,然后函数的参数是基于实际参数的副本最初值,调用,也得到该函数返回的结束值复印件. 请看下面的代码: class Person { publi ...
- C++习题 复数类--重载运算符+
Description 定义一个复数类Complex,重载运算符"+",使之能用于复数的加法运算.将运算符函数重载为非成员.非友元的普通函数.编写程序,求两个复数之和. Input ...