codeforces55B. Smallest number

最新推荐文章于 2023-10-20 21:39:18 发布

飘摇的尘土

最新推荐文章于 2023-10-20 21:39:18 发布

阅读量871

点赞数

CC 4.0 BY-SA版权

分类专栏：数学文章标签： codeforces55B. Small

本文链接：https://blog.youkuaiyun.com/R1986799047/article/details/51468756

数学专栏收录该内容

111 篇文章

订阅专栏

本文介绍了一个算法问题，目标是在给定四个整数和三个运算符的情况下，通过加法或乘法运算来最小化最终得到的单一数值。文章提供了一段C++代码实现，通过全排列的方式尝试所有可能的操作顺序，并选取最小的结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Recently, Vladimir got bad mark in algebra again. To avoid such unpleasant events in future he decided to train his arithmetic skills. He wrote four integer numbers a, b, c, d on the blackboard. During each of the next three minutes he took two numbers from the blackboard (not necessarily adjacent) and replaced them with their sum or their product. In the end he got one number. Unfortunately, due to the awful memory he forgot that number, but he remembers four original numbers, sequence of the operations and his surprise because of the very small result. Help Vladimir remember the forgotten number: find the smallest number that can be obtained from the original numbers by the given sequence of operations.

Input

First line contains four integers separated by space: 0 ≤ a, b, c, d ≤ 1000 — the original numbers. Second line contains three signs ('+' or '*' each) separated by space — the sequence of the operations in the order of performing. ('+' stands for addition, '*' — multiplication)

Output

Output one integer number — the minimal result which can be obtained.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

Examples

input
1 1 1 1
+ + *

output
3

input
2 2 2 2
* * +

output
8

input
1 2 3 4
* + +

output
9

题意：给出4个正数和3操作符每次可以从四个数中选一个按照操作符所给的顺序运算再从剩下的数中在选两个运算求最后所剩的一个数最小是多少

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<list>
#include<queue>
#include<cmath>
#include<vector>
using namespace std;
char oper[5];
long long num[5],ans=1000000000000000ll;
long long calu(long long a,long long b,char o){
	if(o=='+')return a+b;
	if(o=='*')return a*b;
}
int main()
{
	scanf("%lld%lld%lld%lld",&num[0],&num[1],&num[2],&num[3]);
	getchar();
	scanf("%c %c %c",&oper[0],&oper[1],&oper[2]);
	int id[5];id[0]=0;id[1]=1;id[2]=2;id[3]=3;
	do
	{
		long long a=calu(num[id[0]],num[id[1]],oper[0]);
		long long b=calu(a,num[id[2]],oper[1]);
		long long temp=calu(b,num[id[3]],oper[2]);
		a=calu(num[id[0]],num[id[1]],oper[0]);
		b=calu(num[id[2]],num[id[3]],oper[1]);
		long long temp1=calu(a,b,oper[2]);
		ans=min(ans,min(temp,temp1));
	}while(next_permutation(id,id+4));
	printf("%lld\n",ans);
	return 0;
}