POJ_2065 SETI 【同余高斯消元】

一、题目

　　 SETI

二、分析　　

　　给定一个模数，一串字符串，字符串长度为N,相当于是N个方程的答案，而这N个方程中有N个未知数，要求的就是这N个未知数的值，很显然的高斯消元，遇到模数和除法，用逆元就好。

三、AC代码

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

using namespace std;
#define ll long long
#define Min(a,b) ((a)>(b)?(b):(a))
#define Max(a,b) ((a)>(b)?(a):(b))
const int maxn = 1e2;
int p;
char s[maxn];
int a[maxn][maxn], x[maxn];
int n, equ, var;

int gcd(int a, int b)
{
    return b == 0 ? a : gcd(b, a%b);
}

int lcm(int a, int b)
{
    return a / gcd(a, b) * b;
}

int inv(int a, int m)
{
    if(a == 1)
        return 1;
    return inv(m%a, m) * (m - m/a)%m;
}

int getInt(char c)
{
    if(c == ‘*‘)
        return 0;
    else
        return c - ‘a‘ + 1;
}

int Gauss()
{
    int 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(a[max_r][col] == 0)
        {
            k--;
            continue;
        }
        if(max_r != k)
        {
            for(int i = col; i < (var + 1); i++)
            {
                swap(a[max_r][i], a[k][i]);
            }
        }
        //消元
        for(int i = k + 1; i < equ; i++)
        {
            if(a[i][col] != 0)
            {
                int LCM = lcm( fabs(a[i][col]), fabs(a[k][col]));
                int ta = LCM / fabs(a[i][col]);
                int tb = LCM / fabs(a[k][col]);
                //异号
                if(a[i][col]*a[k][col] < 0)
                    tb = -tb;
                for(int j = col; j < (var + 1); j++)
                {
                    a[i][j] = ((a[i][j] * ta - a[k][j] * tb) % p + p)%p;
                    a[i][col] = 0;
                }
            }
        }
    }
    for(int i = k; i < equ; i++)
        {
            if(a[i][var+1] != 0)
                return -1;  //无解
        }
        //多解
        if(k < var)
            return var - k;
        for(int i = var - 1; i >= 0; i--)
        {
            int tmp = a[i][var];
            for(int j = i + 1; j < var ;j++)
            {
                if(a[i][j] != 0)
                {
                    tmp -= a[i][j] * x[j];
                }
                tmp = (tmp % p + p) % p;
            }
            x[i] = (tmp * inv(a[i][i], p)) % p;
        }
}

void solve()
{
    equ = n, var = n;
    int res = Gauss();
    for(int i = 0; i < n; i++)
    {
        if(i)
            printf(" ");
        printf("%d", x[i]);
    }
    printf("\n");
}

int main()
{
    int T;
    // freopen("input.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %s", &p, s);
        n = strlen(s);
        memset(a, 0, sizeof(a));
        for(int i = 0; i < n; i++)
        {
            int res = 1;
            for(int j = 0; j < n; j++)
            {
                a[i][j] = res;
                res = res * (i + 1) % p;
            }
            a[i][n] = getInt(s[i]);
        }
        solve();
    }
    return 0;
}

原文地址：https://www.cnblogs.com/dybala21/p/11348791.html