Codeforces 513C Second price auction 数学公式求期望

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题目大意：

就是n个公司(n <= 5个)分别自己出价, 每个公司出的价格随即在[Li, Ri]之间的任意一个整数(1 <= Li <= Ri <= 10000), 问随机出来的5个数中第二大的数的期望是多少

大致思路：

纯数学推理做的...有点厉害的样子...

数学公式见代码注释

代码如下：

Result : Accepted Memory : 4 KB Time : 15 ms

/* * Author: Gatevin * Created Time:  2015/2/26 10:22:50 * File Name: poi~.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;/* * E = ∑b*P(b) * P(b)表示第二大的数是b的概率 * P(b)的计算公式如下, 参考Codeforces第三种解法的计算公式(官方那个我看的时候有点问题) * (这里用II表示连乘, ∑表示累和) * P(b) = P(max = b, second = b) + P(max > b, second = b) * P(max = b, second = b) = P(max <= b) - P(max < b) - P(max = b, second < b) *                        = IIP(ri <= b) - IIP(ri < b) - ∑P(ri = b)*IIP(rj < b) (j != i) * P(max > b, second = b) = ∑P(ri > b)*(IIP(rj <= b) - IIP(rj < b)) (j != i) */int n, L[6], R[6];double Less(int j, int i)//return P(rj <= i){    if(i < L[j]) return 0;    if(i >= R[j]) return 1.;    else return (i - L[j] + 1.)/(R[j] - L[j] + 1.);}double equal(int j, int i)//return P(rj == i){    if(i < L[j] || i > R[j]) return 0;    else return 1./(R[j] - L[j] + 1.);}double more(int j, int i)//return P(rj > i){    if(i >= R[j]) return 0;    if(i < L[j]) return 1.;    return (R[j] - i)/(R[j] - L[j] + 1.);}int main(){    scanf("%d", &n);    int ml = 10010, mr = 0;    double ans = 0;    for(int i = 1; i <= n; i++)    {        scanf("%d %d", L + i, R + i);        ml = min(ml, L[i]);        mr = max(mr, R[i]);    }    for(int i = ml; i <= mr; i++)    {        double p1 = 0;        double p11 = 1, p12 = 1, p13 = 0;        for(int j = 1; j <= n; j++)//这些O(n*n)的部分其实可以预处理降低复杂度        {            p11 *= Less(j, i), p12 *= Less(j, i - 1);//IIP(ri <= b) 和 IIP(ri < b)            double tmp = equal(j, i);            for(int k = 1; k <= n; k++)                if(k != j)                    tmp *= Less(k, i - 1);            p13 += tmp;// ∑P(ri = b)*IIP(rj < b)        }        p1 = p11 - p12 - p13;//P(max = b, second = b)        double p2 = 0;        for(int j = 1; j <= n; j++)        {            double tmp1 = more(j, i);            double tmp2 = 1., tmp3 = 1.;            for(int k = 1; k <= n; k++)                if(k != j)                    tmp2 *= Less(k, i), tmp3 *= Less(k, i - 1);            p2 += tmp1*(tmp2 - tmp3);//P(max > b, second = b) = ∑P(ri > b)*(IIP(rj <= b) - IIP(rj < b))        }        ans += (p1 + p2)*i;    }    printf("%.10f\n", ans);    return 0;}

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