luogu3389 【模板】高斯消元法

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int n, x;
double a[105][105];
bool gauss(){
    for(int i=1; i<=n; i++){
        int maxi=i;
        for(int j=i+1; j<=n; j++)
            if(fabs(a[j][i])>fabs(a[maxi][i]))
                maxi = j;
        if(fabs(a[maxi][i])<1e-7)	return false;
        double now=a[maxi][i];
        swap(a[i], a[maxi]);
        for(int j=i; j<=n+1; j++)
            a[i][j] /= now;
        for(int j=1; j<=n; j++)
            if(j!=i){
                now = a[j][i];
                for(int k=i; k<=n+1; k++)
                    a[j][k] -= a[i][k] * now;
            }//一步到位，上下全消
    }
    return true;
}
int main(){
    cin>>n;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n+1; j++){
            scanf("%d", &x);
            a[i][j] = x;
        }
    if(!gauss())	printf("No Solution
");
    else
        for(int i=1; i<=n; i++)
            printf("%.2lf
", a[i][n+1]);
    return 0;
}

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int n, x;
double a[105][105];
bool gauss(){
    for(int i=1; i<=n; i++){
        int maxi=i;
        for(int j=i+1; j<=n; j++)
            if(fabs(a[j][i])>fabs(a[maxi][i]))
                maxi = j;
        if(fabs(a[maxi][i])<1e-7)	return false;
        double now=a[maxi][i];
        swap(a[i], a[maxi]);
        for(int j=i; j<=n+1; j++)
            a[i][j] /= now;
        for(int j=i+1; j<=n; j++){//对下面几行消元构成上三角
                now = a[j][i];
                for(int k=i; k<=n+1; k++)
                    a[j][k] -= a[i][k] * now;
            }
    }
	for(int i=n; i>=1; i--){//回代
		for(int j=1; j<=i-1; j++){
			a[j][n+1] -= a[j][i] * a[i][n+1];
			a[j][i] = 0;
		}
	}
    return true;
}
int main(){
    cin>>n;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n+1; j++){
            scanf("%d", &x);
            a[i][j] = x;
        }
    if(!gauss())	printf("No Solution
");
    else
        for(int i=1; i<=n; i++)
            printf("%.2lf
", a[i][n+1]);
    return 0;
}