7/28 Gauss elimination to solve linear equations + Gauss elimination to solve XOR linear equations + find the combination number II

1. Gauss elimination linear equations

Enter a include  nn  An equation  nn  A system of linear equations with unknown numbers .

The coefficients in the equations are real numbers .

Solving this system of equations .

The following figure shows an example that contains  mm  An equation  nn  An example of a system of linear equations with unknown numbers ：

Input format

The first line contains integers  nn.

Next  nn  That's ok , Each row contains  n+1n+1  A real number , That represents an equation  nn  A coefficient and a constant to the right of the equal sign .

Output format

If a given system of linear equations has a unique solution , A total of  nn  That's ok , Among them the first  ii  Line output the  ii  The solution of an unknown number , Two decimal places are reserved for the result .

If there are countless solutions to a given system of linear equations , The output  Infinite group solutions.

If a given system of linear equations has no solution , The output  No solution.

Data range

1≤n≤1001≤n≤100,
All input coefficients and constants have two decimal places , The absolute values do not exceed  100100.

 

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 110;
const double eps = 1e-8;

int n;
double a[N][N];

int gauss()  //  Gauss elimination , The answer lies in a[i][n] in ,0 <= i < n
{
    int c, r;
    for (c = 0, r = 0; c < n; c ++ )
    {
        int t = r;
        for (int i = r; i < n; i ++ )  //  Find the row with the largest absolute value 
            if (fabs(a[i][c]) > fabs(a[t][c]))
                t = i;

        if (fabs(a[t][c]) < eps) continue;

        for (int i = c; i <= n; i ++ ) swap(a[t][i], a[r][i]);  //  Change the line with the largest absolute value to the top 
        for (int i = n; i >= c; i -- ) a[r][i] /= a[r][c];  //  Change the first place of the current line to 1
        for (int i = r + 1; i < n; i ++ )  //  Use the current row to eliminate all the following columns into 0
            if (fabs(a[i][c]) > eps)
                for (int j = n; j >= c; j -- )
                    a[i][j] -= a[r][j] * a[i][c];

        r ++ ;
    }

    if (r < n)
    {
        for (int i = r; i < n; i ++ )
            if (fabs(a[i][n]) > eps)
                return 2; //  unsolvable 
        return 1; //  There are infinite sets of solutions 
    }

    for (int i = n - 1; i >= 0; i -- )
        for (int j = i + 1; j < n; j ++ )
            a[i][n] -= a[i][j] * a[j][n];

    return 0; //  There is a unique solution 
}


int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < n + 1; j ++ )
            scanf("%lf", &a[i][j]);

    int t = gauss();
    if (t == 2) puts("No solution");
    else if (t == 1) puts("Infinite group solutions");
    else
    {
        for (int i = 0; i < n; i ++ )
        {
            if (fabs(a[i][n]) < eps) a[i][n] = 0;  //  Remove the output  -0.00  The situation of 
            printf("%.2lf\n", a[i][n]);
        }
    }

    return 0;
}

2. Gauss elimination is used to solve XOR linear equations

Similar to linear equations , One line at a time is not 0 Of , Then XOR solution

Enter a include  nn  An equation  nn  A system of XOR linear equations with unknown numbers .

The coefficients and constants in the equations are  00  or  11, The value of each unknown is also  00  or  11.

Solving this system of equations .

Examples of XOR linear equations are as follows ：

M[1][1]x[1] ^ M[1][2]x[2] ^ … ^ M[1][n]x[n] = B[1]
M[2][1]x[1] ^ M[2][2]x[2] ^ … ^ M[2][n]x[n] = B[2]
…
M[n][1]x[1] ^ M[n][2]x[2] ^ … ^ M[n][n]x[n] = B[n]

among  ^  Express exclusive or (XORXOR),M[i][j]M[i][j]  It means the first one  ii  In this equation  x[j]x[j]  The coefficient of ,B[i]B[i]  It's No  ii  The constant at the right end of an equation , All values are  00  or  11.

Input format

The first line contains integers  nn.

Next  nn  That's ok , Each row contains  n+1n+1  It's an integer  00  or  11, That represents an equation  nn  A coefficient and a constant to the right of the equal sign .

Output format

If a given system of linear equations has a unique solution , A total of  nn  That's ok , Among them the first  ii  Line output the  ii  The solution of an unknown number .

If there are multiple solutions for a given system of linear equations , The output  Multiple sets of solutions.

If a given system of linear equations has no solution , The output  No solution.

Data range

1≤n≤1001≤n≤100

sample input ：

3
1 1 0 1
0 1 1 0
1 0 0 1

sample output ：

1
0
0

 

 

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 110;
int n;
int a[N][N];
int gauss()
{
    int c, r;
    for (c = 0, r = 0; c < n; c ++ )
    {
        int t = r;
        for (int i = r; i < n; i ++ )
            if (a[i][c])
                t = i;

        if (!a[t][c]) continue;

        for (int i = c; i <= n; i ++ ) swap(a[r][i], a[t][i]);
        for (int i = r + 1; i < n; i ++ )
            if (a[i][c])
                for (int j = n; j >= c; j -- )
                    a[i][j] ^= a[r][j];
        r ++ ;
    }
    if (r < n)
    {
        for (int i = r; i < n; i ++ )
            if (a[i][n])
                return 2;
        return 1;
    }
    for (int i = n - 1; i >= 0; i -- )
        for (int j = i + 1; j < n; j ++ )
            a[i][n] ^= a[i][j] * a[j][n];
    return 0;
}
int main()
{
    cin >> n;

    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < n + 1; j ++ )
            cin >> a[i][j];
    int t = gauss();
    if (t == 0)
    {
        for (int i = 0; i < n; i ++ ) cout << a[i][n] << endl;
    }
    else if (t == 1) puts("Multiple sets of solutions");
    else puts("No solution");
    return 0;
}

3. Find the combination number  （ Find factorials and inverse elements of factorials ）

Given  nn  Group inquiry , Each group is given two integers  a,ba,b, Please output  Cbamod(109+7)Cabmod(109+7)  Value .

Input format

The first line contains integers  nn.

Next  nn  That's ok , Each line contains a set of  aa  and  bb.

Output format

common  nn  That's ok , Each line outputs a solution to the query .

Data range

1≤n≤100001≤n≤10000,
1≤b≤a≤1051≤b≤a≤105

sample input ：

3
3 1
5 3
2 2

sample output ：

3
10
1

 

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 100010, mod = 1e9 + 7;


int fact[N], infact[N];


int qmi(int a, int k, int p)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}


int main()
{
    fact[0] = infact[0] = 1;
    for (int i = 1; i < N; i ++ )
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;
    }


    int n;
    scanf("%d", &n);
    while (n -- )
    {
        int a, b;
        scanf("%d%d", &a, &b);
        printf("%d\n", (LL)fact[a] * infact[b] % mod * infact[a - b] % mod);
    }

    return 0;
}