摘要：
=j){40for{//在当前矩阵进行变换时,对所求矩阵也进行同样的变换41a[j][k]=%p;42}43a[j][i]=0;44}45}46}47}48intmain(){49scanf;50for{51for{52scanf;53}54a[i][n+i]=1;//另一半初始化为元矩阵55}56m=n˂˂1;57gauss();58for{59for{60printf("%lld%c",a[i][j],j==m?
思路：
求A的逆矩阵，把A和单位矩阵I放在一个矩阵里
对A进行加减消元使A化成单位矩阵
此时原来单位矩阵转化成逆矩阵
原理大概就是 A(逆) * [A I] = [I A(逆)]
Code:
1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespacestd; 4 const double eps = 1E-8; 5 const int p = 1e9 + 7; 6 const int N = 405; 7 intn, m; 8 ll a[N][N << 1]; 9 ll pw(ll x, ll y) { 10 ll re = 1; 11 for (; y; y >>= 1) { 12 if (y & 1) { 13 re = re * x %p; 14 } 15 x = x * x %p; 16 } 17 returnre; 18 } 19 void gauss() {//高斯消元 20 for (int i = 1; i <= n; i++) { 21 int pre =i; 22 for (int j = i + 1; j <= n; j++) { 23 if (fabs(a[j][i] - a[pre][i]) >=eps) { 24 pre =j; 25 } 26 } 27 for (int k = 1; k <= m; k++) {//交换着两行,记得把右侧新添矩阵也交换 28 swap(a[i][k], a[pre][k]); 29 } 30 if (fabs(a[i][i]) <=eps) { 31 printf("No Solution "); 32 exit(0); 33 } 34 ll inv = pw(a[i][i], p - 2);//求逆元 35 for (int k = i; k <= m; k++) { 36 a[i][k] = a[i][k] * inv %p; 37 } 38 for (int j = 1; j <= n; j++) { 39 if (i !=j) { 40 for (int k = i + 1; k <= m; k++) {//在当前矩阵进行变换时,对所求矩阵也进行同样的变换 41 a[j][k] = (a[j][k] - a[j][i] * a[i][k] % p + p) %p; 42 } 43 a[j][i] = 0; 44 } 45 } 46 } 47 } 48 intmain () { 49 scanf("%d", &n); 50 for (int i = 1; i <= n; i++) { 51 for (int j = 1; j <= n; j++) { 52 scanf("%lld", &a[i][j]); 53 } 54 a[i][n + i] = 1;//另一半初始化为元矩阵 55 } 56 m = n << 1; 57 gauss(); 58 for (int i = 1; i <= n; i++) { 59 for (int j = n + 1; j <= m; j++) { 60 printf("%lld%c", a[i][j], j == m ? '' : ' '); 61 } 62 } 63 return 0; 64 }