UVA 684 (矩阵行列式)

来源：互联网发布：江苏省网络协会编辑：程序博客网时间：2024/04/30 13:55

题目链接:点击打开链接

求行列式的值,自己写了个板子,精度的地方有点蛋疼然后就用了很麻烦的方式.额,,,

比较实用的.

行列式两行交换结果是要*-1的.

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <vector>#include <queue>#include <cmath>using namespace std;#define maxn 33#define eps 1e-10int n;double a[maxn][maxn];long long solve () {    double ans = 1.0;    int i, j, k, col, max_r;    for (k = 0, col = 0; k < n && col < n; k++, col++) {//转化成上三角矩阵        max_r = k;        for (int i = k+1; i < n; i++) { //找到最大的绝对值数所在的行            if (fabs (a[i][col]) > fabs (a[max_r][col]))                max_r = i;        }        if (k != max_r) { //交换行            for (int j = col; j < n; j++) {                swap (a[k][j], a[max_r][j]);            }            ans *= (-1);//行列式换行改变符号        }        for (int i = k+1; i < n; i++) { //消去            if (a[i][col]) {                double tmp = -a[i][col]/a[k][col];                for (int j = col; j < n; j++) {                    a[i][j] += tmp*a[k][j];                }            }        }    }    for (i = 0; i < n; i++) ans *= a[i][i];    long long p1 = ceil (ans), p2 = floor (ans);    if (ans-p2 < p1-ans)        return p2;    else        return p1;}int main () {   //freopen ("in.txt", "r", stdin);    while (1) {        cin >> n;        if (n == 0) {            cout << "*" << "\n";            break;        }        for (int i = 0; i < n; i++) {            for (int j = 0; j < n; j++)                cin >> a[i][j];        }        cout << solve () << "\n";    }    return 0;}

另外扒了一份矩阵辗转相除的板子,可以避免精度误差.

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <vector>#include <queue>using namespace std;#define maxn 33#define eps 1e-10long long a[maxn][maxn];int n;long long det ()        //按列化为下三角{     long long res = 1;     for(int i = 0; i < n; ++i)     {         if (!a[i][i])         {             bool flag = false;             for (int j = i + 1; j < n; ++j)             {                 if (a[j][i])                 {                     flag = true;                     for (int k = i; k < n; ++k)                     {                         swap (a[i][k], a[j][k]);                     }                     res = -res;                     break;                 }             }             if (!flag)             {                 return 0;             }         }         for (int j = i + 1; j < n; ++j)         {             while (a[j][i])             {                 long long t = a[i][i] / a[j][i];                 for (int k = i; k < n; ++k)                 {                     a[i][k] = a[i][k] - t * a[j][k];                     swap (a[i][k], a[j][k]);                 }                 res = -res;             }          }          res *= a[i][i];      }      return res;}int main(){    //freopen("in.txt", "r", stdin);    while (1) {        cin >> n;        if (n == 0) {            cout << "*\n";            break;        }        for (int i = 0; i < n; i++) {            for (int j = 0; j < n; j++)                cin >> a[i][j];        }        cout << det () << "\n";    }    return 0;}

0 0