高斯消元法

高斯消元法求解异或方程;

poj1222模板题，网上有许多此题的题解，但还是觉得这个讲的比较好懂 http://blog.youkuaiyun.com/zhuichao001/article/details/5440843

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<cctype>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<string>
#include<set>
#include<stack>
#define ll long long
#define MAX 1000
#define INF INT_MAX
#define eps 1e-8

using namespace std;

int a[MAX][MAX];

void gauss(){
	for (int i=0; i<30; i++){
		int k;
		for (int j=i; j<30; j++) if(a[j][i]){
			k = j;
			break;
		} 
		if (k != i){
			for (int j=0; j<=30; j++) swap(a[i][j],a[k][j]);
		}
		for (int j=0; j<30; j++){
			if (i != j && a[j][i]){
				for (int u=0; u<=30; u++){
					a[j][u] ^= a[i][u];
				}
			}
		}
	}
}

int main(){
	int T;
	scanf("%d",&T);
	for (int cas = 1; cas <= T; cas++){
		for (int i=0; i<30; i++){
			scanf("%d",&a[i][30]);
		}
		for (int i=0; i<30; i++){
			a[i][i] = 1;
			if (i % 6 == 0){
				a[i+1][i] = 1;
			}
			else if (i % 6 == 5){
				a[i-1][i] = 1;
			}
			else {
				a[i-1][i] = 1;
				a[i+1][i] = 1;
			}
			if (i / 6 == 0){
				a[i+6][i] = 1;
			}
			else if (i / 6 == 4){
				a[i-6][i] = 1;
			}
			else {
				a[i+6][i] = 1;
				a[i-6][i] = 1;
			}
		}
		/*for (int i=0; i<30; i++){
			for (int j=0; j<30; j++) printf("%d ",a[i][j]);
			printf("\n");
		}*/
		gauss();
		printf("PUZZLE #%d\n",cas);
		for (int i=0; i<30; i++){
			printf("%d",a[i][30]);
			if (i%6 == 5) printf("\n");
			else printf(" ");
		}
	}
	return 0;
}

高斯消元法求解的个数，高精度模板；

fzu1704

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<climits>
#include<cctype>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<string>
#define ll long long
#define MAX 1000
#define eps 1e-6
#define INF INT_MAX

using namespace std;

int a[MAX][MAX],n,m;

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

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

void two_power(int x){
	int str[MAX];
	memset(str,0,sizeof(str));
	str[0] = 1;
	int c = 1;
	for (int i=1; i<=x; i++){
		int g = 0;
		for (int j=0; j<c; j++){
			str[j] = str[j]*2 + g;
			g = str[j] / 10;
			str[j] = str[j] % 10;
		}
		if (g != 0) str[c++] = g;
	}
	for (int i = c-1; i>=0; i--) printf("%d",str[i]); printf("\n");
	return;
}
 
void Gauss()  
{  
    int  u, v;  
    for(u=0,v=0; u<n&&v<m; u++,v++)  
    {  
        int r = u;  
        for(int i=u+1; i<n; i++)  
            if(a[i][v]>a[r][v]) r = i;  
        if(r != u)  
            for(int i=u; i<=m; i++)  swap(a[u][i],a[r][i]);  
        if(a[u][v]==0) {
		 	u--; 
			continue; 
		}  
        for(int i=u+1; i<n; i++)  
           if(a[i][v] != 0)  
           {  
               int lcm=LCM(a[i][v], a[u][v]);  
               int ta=lcm / a[i][v], tb=lcm / a[u][v];  
               if(a[i][v] * a[u][v] < 0)  tb = -tb;  
               for(int j=v; j<=m; j++)  
                 a[i][j]=((a[i][j]*ta)%2 - (a[u][j]*tb)%2+2)%2;  
           }  
     }  
     for(int i=u; i<n; i++)  
        if(a[i][v]!=0) {puts("0");return;}  
    two_power(m - u);  
}

int main(){
	int T ;
	scanf("%d",&T);
	while (T--){
		scanf("%d%d",&n,&m);
		memset(a,0,sizeof(a));
		for (int i=0; i<n; i++) scanf("%d",&a[i][m]);
		for (int i=0; i<m; i++){
			int u,v;
			scanf("%d",&u);
			while (u--){
				scanf("%d",&v);
				v--;
				a[v][i] = 1;
			}
		}
		Gauss();
	}
	return 0;
}

hdu2449高精度高斯消元模板求方程组的解，要求结果为分数形式；

http://www.cnblogs.com/DrunBee/archive/2012/09/04/2669797.html

在模P的线性空间下的高斯消元法求解方程组的 模板题目poj2065

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<climits>
#include<cctype>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<string>
#define ll long long
#define MAX 100
#define eps 1e-6
#define INF INT_MAX

using namespace std;

int a[MAX][MAX],x[MAX],n,m;

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

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

int mod_mul(int a, int b, int N)  
{  
    int ret = 0;  
    while (b)  
    {  
        if (b & 1)  
            ret = (ret + a) % N;  
        a = 2 * a % N;  
        b >>= 1;  
    }  
    return ret;  
}  

int mod_pow(int x, int n, int N)      
{  
    int ret = 1;  
    x %= N;  
    while (n)  
    {  
        if (n & 1)  
            ret = mod_mul(ret, x, N);  
        x = mod_mul(x, x, N);  
        n >>= 1;  
    }  
    return ret;  
}  

void Gauss(int P)             //模P下的高斯消元法解方程组 
{  
    int u, v;  
    for(u=0,v=0; u<n&&v<m; u++,v++)  
    {  
        int r = u;  
        for(int i=u+1; i<n; i++)  
            if(a[i][v]>a[r][v]) r = i;  
        if(r != u)  
            for(int i=u; i<=m; i++)  swap(a[u][i],a[r][i]);  
        if(a[u][v] == 0) {
		 	u--; 
			continue; 
		}  
        for(int i=u+1; i<n; i++)  
           if(a[i][v] != 0)  
           {  
               int lcm=LCM(a[i][v], a[u][v]);  
               int ta=lcm / a[i][v], tb=lcm / a[u][v];  
               if(a[i][v] * a[u][v] < 0)  tb = -tb;  
               for(int j=v; j<=m; j++)  
                 a[i][j] = ((a[i][j]*ta) % P - (a[u][j]*tb) % P + P) % P;  
           }  
     }  
     
     for (int i = u; i < m; i++) if (a[i][m] != 0) return;   //有解性判断
	 
	 if (u < m) return;           //无穷解的情况,其中m-u为自由变元的数量 
     
     for (int i=m-1; i>=0; i--){      //回带求解过程 
     	int tmp = a[i][m];
     	for (int j=i+1; j<m; j++){
     		tmp = ((tmp - a[i][j]*x[j]) % P + P ) % P;
     	}
     	while (tmp % a[i][i] != 0){      //将解最好化为整数（模P下有解则一定存在整数解） 
     		tmp += P;
     	}
     	x[i] = (tmp / a[i][i]) % P;
     }
}

int main(){
	int T,P;
	char s[MAX];
	scanf("%d",&T);
    while (T--){
		scanf("%d%s",&P,s);
		n = m = strlen(s);
		for (int i=0; i<n; i++){
		     if (s[i] == '*')
		     	a[i][m] = 0;
		     else 
		     	a[i][m] = s[i] - 'a' + 1;
		}
		for (int i=0; i<n; i++){
			for (int j=0; j<m; j++)
				a[i][j] = mod_pow(i+1,j,P);
		}
		Gauss(P);
		for (int i=0; i<n-1; i++) printf("%d ",x[i]);
		printf("%d\n",x[n-1]);
	}
	return 0;
}