uva10700- Camel trading

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.
题意：
如上，给你一个带‘+’和‘*’的表达式，求最大和最小值
思路：
最大值：先加后乘
最小值：先乘后加
因为 ： （a+b）*c a+b*c
->a*c+b*c a+b*c
都是正整数，显然先加后乘更大。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1005;
char str[N];
long long  num[N];
char op[N];
int len, lop, lnum;

void init() {
    lop = 0, lnum = 0;
    for (int i = 0; i < len;) {
        if (str[i] == '+' || str[i] == '*') {
            op[lop++] = str[i];
            i++;
        }
        else {
            long long  number = 0;
            while (str[i] >= '0'&&str[i] <= '9') {
                number = number * 10 + (str[i] - '0');
                i++;
            }
            num[lnum++] = number;
        }
    }
    /*for (int i = 0; i < lop; i++)
    printf("%c\t", op[i]);
    printf("\n");
    for (int i = 0; i < lnum; i++)
    printf("%d\t", num[i]);
    printf("\n");*/
}
long long  get_min() {
    long long  temp[N];
    long long  minn = 0;
    for (int i = 0; i < lnum; i++)
        temp[i] = num[i];
    for (int i = 0; i < lop; i++) {
        if (op[i] == '*') {
            temp[i + 1] *= temp[i];
            temp[i] = 0;
        }
    }
    for (int i = 0; i < lnum; i++) {
        minn += temp[i];
    }
    return minn;
}
long long  get_max() {
    long long temp[N];
    long long maxn = 1;
    for (int i = 0; i < lnum; i++)
        temp[i] = num[i];
    for (int i = 0; i < lop; i++) {
        if (op[i] == '+') {
            temp[i + 1] += temp[i];
            temp[i] = 0;
        }
    }
    for (int i = 0; i < lnum; i++) {
        if (temp[i] != 0)
            maxn *= temp[i];
    }
    return maxn;
}
int main() {
    int cas;
    scanf("%d\n", &cas);
    while (cas--) {
        gets(str);
        len = strlen(str);
        init();
        long long minn = get_min();
        long long maxn = get_max();
        printf("The maximum and minimum are %lld and %lld.\n", maxn, minn);
    }

    return 0;
}