UVA - 10700

Camel trading

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Description

Problem E - Camel trading

Time Limit: 1 second

Background

Aroud 800 A.D., El Mamum, Calif of Baghdad was presented the formula 1+2*3*4+5, which had its origin in the financial accounts of a camel transaction. The formula lacked parenthesis and was ambiguous. So, he
decided to ask savants to provide him with a method to find which interpretation is the most advantageous for him, depending on whether is is buying or selling the camels.

The Problem

You are commissioned by El Mamum to write a program that determines the maximum and minimum possible interpretation of a parenthesis-less expression.

Input

The input consists of an integer N, followed by N lines, each containing an expression. Each expression is composed of at most 12 numbers, each ranging between 1 and 20,
and separated by the sum and product operators +and *.

Output

For each given expression, the output will echo a line with the corresponding maximal and minimal interpretations, following the format given in the sample output.

Sample input

3
1+2*3*4+5
4*18+14+7*10
3+11+4*1*13*12*8+3*3+8

Sample output

The maximum and minimum are 81 and 30.
The maximum and minimum are 1560 and 156.
The maximum and minimum are 339768 and 5023.

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy :: Non
Classical, Usually Harder

Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design

Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy
- Standard

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思路：很明显，当加法优于乘法先算出时有最大，乘法先算出时有最小

AC代码（略挫，没用stack写）：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define LL long long
using namespace std;

int N;
char str[100];
LL ope[20];
LL num[20];
LL sum, osum;

LL fun(char s[], int n) {	//字符串转换为数字
	LL tmp = 0, cnt = 0;
	for(int i = n; isdigit(s[i]); i++) {
		tmp = tmp * 10 + (s[i] - '0');
		cnt ++;
	}
	num[sum++] = tmp;
	return cnt - 1;
}

int main() {
	scanf("%d", &N);
	while(N--) {
		sum = osum = 0;
		memset(num, 0, sizeof(num));
		memset(ope, 0, sizeof(ope));
		scanf("%s", str);
		for(int i = 0; str[i] != '\0'; i++) {
			if(isdigit(str[i])) {
				i += fun(str, i);
			}
			else
			{
				if(str[i] == '*') ope[osum++] = 0;
				else ope[osum++] = 1;
			}
		}

		//for(int i = 0; i < sum; i++) printf("%d ", num[i]);
		//printf("\n");
		//for(int i = 0; i < osum; i++) printf("%d ", ope[i]);
		//printf("\n");

		LL MAX = 1, MIN = 0;
		//找最小
		LL tnum[20];
		for(int i = 0; i < 20; i++) tnum[i] = num[i];
		for(int i = 0; i < osum; i++) {
			if(ope[i] == 0) {
				int t = i;
				LL tmp = tnum[i] * tnum[i + 1];
				tnum[i + 1] = 0;
				for(int j = i + 1; j < osum && ope[j] == 0; j++) {
					tmp *= tnum[j + 1];
					i = j;
					tnum[j + 1] = 0;
				}
				tnum[t] = tmp;
			}
		}
		//for(int i= 0; i < sum; i++) printf("%d ", tnum[i]);
		for(int i = 0; i < sum; i++) MIN += tnum[i];
		//找最大
		for(int i = 0; i < osum; i++) {
			if(ope[i] == 1) {
				int t = i;
				LL tmp = num[i] + num[i + 1];
				num[i + 1] = 1;
				for(int j = i + 1; j < osum && ope[j] == 1; j++) {
					tmp += num[j + 1];
					i = j;
					num[j + 1] = 1;
				}
				num[t] = tmp;
			}
		}
		for(int i = 0; i < sum; i++) MAX *= num[i];

		printf("The maximum and minimum are %lld and %lld.\n", MAX, MIN);
	}
	return 0;
}