UVA 10700 Camel trading 无括号的表达式 贪心

题意：给出只包含数字和+*的表达式，你可以自己安排每一个运算的顺序，让你找出表达式可能得到的最大值和最小值。

   很明显，先乘后加是最小值，先加后乘能得到最大值。

   其实不是很明显。。。

证明下：

   数字的范围是大于等于1的，所以a+b*c如果先考虑加法就变成(a+b)*c，变换下就是ac+bc，而先考虑乘法的话就是a+bc，由于c是大于等于1的，所以ac+bc>=a+bc，先考虑加法结果会不小于先考虑乘法。

根据这个思想，先乘后加是最小值，先加后乘能得到最大值。

代码：

 /* *   Author:        illuz <[e-mail]> *   Blog:          http://blog.csdn.net/hcbbt *   File:          uva10700.cpp *   Lauguage:      C/C++ *   Create Date:   2013-08-25 16:58:52 *   Descripton:    UVA 10700 Camel trading, simulation */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <iostream>#include <list>#include <vector>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <utility>#include <algorithm>using namespace std;#define rep(i, n) for (int i = 0; i < (n); i++)#define repu(i, a, b) for (int i = (a); i < (b); i++)#define repf(i, a, b) for (int i = (a); i <= (b); i++)#define repd(i, a, b) for (int i = (a); i >= (b); i--)#define swap(a, b) {int t = a; a = b; b = t;}#define mc(a) memset(a, 0, sizeof(a))#define ms(a, i) memset(a, i, sizeof(a))#define sqr(x) ((x) * (x))#define FI(i, x) for (typeof((x).begin()) i = (x).begin(); i != (x).end(); i++)typedef long long LL;typedef unsigned long long ULL;/****** TEMPLATE ENDS ******/const int MAXN = 30;char o[MAXN];LL s[MAXN], m;LL Min() {LL sum = 0;rep(i, m) {LL t = s[i];while(o[i] == '*') t *= s[++i];sum += t;}return sum;}LL Max() {LL sum = 1;rep(i, m) {LL t = s[i];while(o[i] == '+') t += s[++i];sum *= t;}return sum;}int main() {int n;scanf("%d\n", &n);while (n--) {m = -1;mc(o); mc(s);while (o[m++] != '\n')scanf("%lld%c", &s[m], &o[m]);printf("The maximum and minimum are %lld and %lld.\n", Max(), Min());}return 0;}