UVA - 11291 Smeech （模拟）

Description

Problem B: Smeech

Professor Octastichs has invented a new programming language, Smeech. An expression in Smeech may be a positive or negative integer, or may be of the form(pe₁e₂) where p is a real number between 0 and 1 (inclusive) ande₁ and e₂ are Smeech expressions. The value represented by a Smeech expression is as follows:

• An integer represents itself
• With probability p, (pe₁e₂) representsx+y where x is the value of e₁ and y is the value ofe₂; otherwise it represents x-y.

Given a Smeech expression, what is its expected value?

Input consists of several Smeech expressions, one per line, followed by a line containing (). For each expression, output its expected value to two decimal places.

Sample Input

7(.5 3 9)()

Output for Sample Input

7.003.00题意：给出个表达式(p,e1,e2)表示(e1+e2)*p+(e1-e2)*(1-p)的结果，求最后的值思路：递归的处理整个式子，注意细节小数的时候的判断
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 10005;char str[maxn];int cur, len;double cal() {double op = 1;while (!(str[cur]=='(' || (str[cur]>='0'&&str[cur]<='9') || str[cur]=='-') && cur < len)cur++;if (str[cur] == '-') {cur++;op = -1;}if (str[cur] == '(') {cur++;double w = 0.1, p = 0;if (str[cur] == '.' || str[cur+1] == '.') {if (str[cur+1] == '.')cur++;for (int i = cur+1; str[i] >= '0' && str[i] <= '9' && str[i]; i++, cur = i) {p = p + (str[i] - '0') * w;w *= 0.1;}}else p = str[cur++] - '0';double a, b;a = cal();b = cal();return (a + b) * p + (a - b) * (1 - p);}else {double p = 0;for (int i = cur; str[i] >= '0' && str[i] <= '9' && str[i]; i++, cur = i)p = p * 10 + str[i] - '0';return op * p;}}int main() {double p, a, b;while (gets(str) && strcmp(str, "()")) {len = strlen(str);cur = 0;printf("%.2lf\n", cal());}return 0;}