高斯消元法

原创于 2017-10-21 15:45:14 发布 · 219 阅读

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#高斯消元

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本文介绍了使用高斯消元法解决线性方程组的方法，并提供了一个具体的C++实现示例。此外，还探讨了如何利用费马小定理进行矩阵运算，包括矩阵求逆等操作。

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Describe

对于一个n个变量的n个方程的式子,惊醒求解!
这里写图片描述
经过了…

变成了…
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最后就是…
这里写图片描述

代码(普通高斯消元)

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#define ll long long
#define N 205
using namespace std;
int n;
double a[N][N],val[N],score[N];
void Gossi()
{
    for(int i=1; i<=n; i++) // ²Ù×÷µÄµÚ¼¸¸ö Êý×Ö ±ÈÈç 1 2 3 4 ...
        for(int j=i+1; j<=n; j++) //  Ã¶¾ÙÏÂÃæµÄÁÐ i+1 i+2 i+3...
        {
            int loc=i+1;
            while(!a[i][i]&&loc<=n)
            {
                swap(a[i],a[loc]);
                loc++;
            }
            double res=a[i][i]/a[j][i];
            for(int v=i+1; v<=n; v++)
                a[j][v]=(a[j][v]*res)-a[i][v];
            score[j]=res*score[j]-score[i];
        }
    val[n]=score[n]/a[n][n];
    for(int i=n-1; i>=1; i--)
    {
        double res=0;
        for(int j=i+1; j<=n; j++)
            res+=val[j]*a[i][j];
        val[i]=(score[i]-res)/a[i][i];
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
            scanf("%lf",&a[i][j]);
        scanf("%lf",&score[i]);
    }
    Gossi();
    for(int i=1; i<=n; i++)
        printf("%.2lf\n",val[i]);
    return 0;
}

费马小定理(Gauss)

#include <bits/stdc++.h>
#define LL long long
#define int long long
using namespace std;
const int m1 = (1e9 + 7);
const int m2 = (1e9 + 9);
const int maxn = 210;
LL m,temp,a[maxn][maxn],b[maxn][maxn],A[maxn],B[maxn];
int n; char s;
LL po(LL x,int y,int mo)
{
    if (y == 1) return x;
    LL res = po(x,y >> 1,mo);
    res = (res * res) % mo;
    if (y & 1) res = (res * x) % mo;
    return res;
}
signed main()
{
    scanf("%d",&n);
    for (int i = 1;i <= n;++i)
    {
        for (int j = 1;j <= n;++j) scanf("%lld",&a[i][j]),b[i][j]=a[i][j];
        LL x1 = 0, x2 = 0;
        while ( (s = getchar())&&!isdigit(s));
        while (isdigit(s))
        {
            x1 = (x1 * 10 + s - '0')%m1;
            x2 = (x2 * 10 + s - '0')%m2;
            s = getchar();
        }
        a[i][0]=m1 - x1; b[i][0]=m2 - x2;
    }
    for (int i = 1;i < n;++i)
    {
        LL x=po(a[i][i],m1-2,m1);
        a[i][0]=a[i][0]*x%m1;
        for (int j = i;j <= n;++j) a[i][j]=a[i][j]*x%m1;
        for (int j = i + 1;j <= n;++j)
        {
            x=a[j][i]; a[j][0] = (a[j][0]-x*a[i][0]%m1)%m1;
            for (int k = i;k <= n;++k) a[j][k]=(a[j][k]-x*a[i][k]%m1)%m1;
        }
    }
    for (int i = n;~i;--i)
    {
        A[i]=-a[i][0] % m1 * po(a[i][i],m1-2,m1)%m1;
        for (int j = 1;j < i;++j) a[j][0]=(a[j][0]+a[j][i]*A[i]%m1)%m1;
    }
    for (int i = 1;i < n;++i)
    {
        int x=po(b[i][i],m2-2,m2); b[i][0]=(b[i][0]*x + m2)%m2;
        for (int j = i;j <= n;++j) b[i][j]=(b[i][j]*x + m2)%m2;
        for (int j = i + 1;j <= n;++j){
            x = b[j][i]; b[j][0] = (b[j][0]-x*b[i][0]%m2 + m2)%m2;
            for (int k = i;k <= n;++k) b[j][k]=(b[j][k] - x * b[i][k]%m2 + m2)%m2;
        }
    }
    for (int i = n;i > 0;--i){
        B[i] = (-b[i][0])*po(b[i][i],m2-2,m2)%m2;
        for (int j=1;j<i;++j) b[j][0]=(b[j][0]+b[j][i]*B[i]%m2)%m2;
    }
    int x = (m1 + 1) / 2; m=1ll*m1*m2;
    for (int i=1;i<=n;++i)
    {
        temp=(A[i]*m2+B[i]*m1)%m;
        if (temp<0) temp+=m;
        printf("%lld\n",(temp*x-(LL)((double)temp/m*x)* m + m)%m);
    }
    return 0;
}