Codeforces Beta Round #51 B. Smallest number dfs

B. Smallest number

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/55/problem/B

Description

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).

Sample Input

1 1 1 1
+ + *

Sample Output

3

HINT

题意

给你4个数，然后给你三个符号，问你怎么安排着4个数的顺序，可以使得最后的答案最小

4个数和三个符号，选两个数和第一个符号出来，得到一个数

3个数和2个符号，选两个数和第二个符号出来，得到一个数

2个数和1个符号，得到答案

要求使得答案最小

题解：

暴力枚举。。。

代码：

#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;

long long ans = 1LL<<61;
char S[10];
void solve2(long long a,long long b) {
    long long x = (S[4] == '+') ? (a + b) : (a * b);
    if (ans > x) {
        ans = x;
    }
}
void solve3(long long a,long long b,long long c) {
    if (S[3] == '+') {
        solve2(a + b, c);
        solve2(a + c, b);
        solve2(b + c, a);
    } else {
        solve2(a * b, c);
        solve2(a * c, b);
        solve2(b * c, a);
    }
}
void solve4(long long a,long long b,long long c,long long d) {
    if (S[2] == '+') {
        solve3(a + b, c, d);
        solve3(a + c, b, d);
        solve3(a + d, b, c);
        solve3(b + c, a, d);
        solve3(b + d, a, c);
        solve3(c + d, a, b);
    } else {
        solve3(a * b, c, d);
        solve3(a * c, b, d);
        solve3(a * d, b, c);
        solve3(b * c, a, d);
        solve3(b * d, a, c);
        solve3(c * d, a, b);
    }
}

int main() {
    long long a, b, c, d;
    cin>>a>>b>>c>>d;
    for(int i=2;i<=4;i++)
        cin>>S[i];
    solve4(a,b,c,d);
    cout<<ans<<endl;
}