UVA - 11291 Smeech （模拟）

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Description

Problem B: Smeech

• An integer represents itself
• With probability p, (pe₁e₂) represents x+y where x is the value of e₁ and y is the value of e₂; otherwise it represents x-y.

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

7.00
3.00
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;
}

UVA - 11291 Smeech （模拟）

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原文地址：http://blog.csdn.net/u011345136/article/details/39083183