高斯消元法

Describe

对于一个n个变量的n个方程的式子,惊醒求解!

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代码(普通高斯消元)

#include <iostream>#include <stdio.h>#include <algorithm>#include <cstring>#define ll long long#define N 205using 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++) //  ö¾ÙÏÂÃæµÄÁÐ 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 longusing 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;}