SDUT 2878 Circle (高斯消元)

题意

初始时站在0点，每次有0.5的几率左移一格或者右移一格，求移动到x格的期望。

思路

显然我们能写出来 Exp[i]=0.5∗Exp[i−1]+0.5∗Exp[i+1]+1
这样的式子我们能写出n个，然后这个式子转化一下变成 Exp[i]−0.5∗Exp[i−1]−0.5∗Exp[i+1]=1 的形式我们就能构造出一个n个变元n个方程的方程组了。

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int maxn = 1e5 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);
int n, xx, equ, var;
double a[1010][1010];
bool free_x[1010];
double x[1010];

/* 返回-1无解 返回0唯一解 大于0多解，且返回自由变元个数 可以用free_x判断不确定的解 */
int gauss()
{
    int i, j, k, col, max_r;
    for (k = 0, col = 0; k < equ && col < var; k++, col++)
    {
        max_r = k;
        for (int i = k + 1; i < equ; i++)
            if (fabs(a[i][col]) > fabs(a[max_r][col]))
                max_r = i;
        if (fabs(a[max_r][col]) < eps) return 0;
        if (k != max_r)
        {
            for (j = k; j <= var; j++)
                swap(a[k][j], a[max_r][j]);
        }
        if (fabs(a[k][col]) == 0)
        {
            k--;
            continue;
        }
        for (i = k + 1; i < equ; i++)
        {
            if (fabs(a[i][col]) == 0)
                continue;
            double temp = a[i][col] / a[k][col];
            for (j = col; j <= var; j++)
                a[i][j] -= a[k][j] * temp;
        }
    }
    //无解
    for (i = k; i < equ; i++)
    {
        if (fabs(a[i][col]) > eps)
            return -1;
    }
    //无穷解
    if (k < var)
    {
        for (i = k - 1; i >= 0; i--)
        {
            int free_num = 0;      //变元个数
            int free_index;        //变元
            for (j = 0; j < n; j++)
            {
                if (fabs(a[i][j]) && free_x[j])
                    free_num++, free_index = j;
            }
            if (free_num > 1)      //有两个以上的变元无法确定
                continue;
            double temp = a[i][n];
            for (j = 0; j < n; j++)
            {
                if (fabs(a[i][j]) && j != free_index)
                    temp -= a[i][j] * x[j];
            }
            x[free_index] = temp / a[i][free_index];
            free_x[free_index] = 0;
        }
        return var - k;
    }
    for (i = equ - 1; i >= 0; i--)
    {
        double temp = a[i][n];
        for (j = i + 1; j < var; j++)
            if (fabs(a[i][j]) != 0)
                temp -= a[i][j] * x[j];
        x[i] = temp / a[i][i];
    }
    return 0;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &xx);
        memset(a, 0, sizeof(a));
        memset(x, 0, sizeof(x));
        for (int i = 0; i < n; i++)
        {
            if (i == xx)
            {
                a[i][i] = 1;
                a[i][n] = 0;
            }
            else
            {
                a[i][i] = 1;
                a[i][n] = 1;
                a[i][(i+n-1)%n] = -0.5;
                a[i][(i+n+1)%n] = -0.5;
            }
        }
        equ = n, var = n;
        gauss();
        printf("%.4f\n", x[0]);
    }
    return 0;
}