SDUT 2878 Circle (高斯消元)

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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;}

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