标签：变换 题目 solution class clu put 相同 模板 long
题目大意：给你一个矩阵$A$，求它的逆矩阵$A^{-1}$，使得$AA^{-1}=I$
题解：设$A=IE_1E_2\cdots E_k$（$E_i$为一个变换），那么$A^{-1}=E_k^{-1}E_{k-1}^{-1}\cdots E_{1}^{-1}$，可以在$A$变为$I$的时候对$I$做相同的操作。当$A$变为$I$时，$I$就变成了$A^{-1}$
卡点：无
C++ Code：
#include <algorithm>
#include <cstdio>
const int mod = 1e9 + 7;
inline int pw(int base, int p) {
static int res;
for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
return res;
}
inline int inv(int x) { return pw(x, mod - 2); }
inline void reduce(int &x) { x += x >> 31 & mod; }
#define maxn 405
int n, nn;
int s[maxn][maxn << 1];
int main() {
scanf("%d", &n); nn = n + n;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) scanf("%d", s[i] + j);
s[i][n + i] = 1;
}
for (int i = 1; i <= n; ++i) {
int pos = i;
while (pos <= n && !s[pos][i]) ++pos;
if (pos > n) {
puts("No Solution");
return 0;
}
for (int j = i; j <= nn; ++j) std::swap(s[pos][j], s[i][j]);
const int Inv = inv(s[i][i]);
for (int j = i; j <= nn; ++j) s[i][j] = static_cast<long long> (s[i][j]) * Inv % mod;
for (int j = 1; j <= n; ++j) if (i != j) {
const int t = s[j][i];
for (int k = i; k <= nn; ++k) reduce(s[j][k] -= static_cast<long long> (s[i][k]) * t % mod);
}
}
for (int i = 1; i <= n; ++i) {
for (int j = n + 1; j <= nn; ++j) printf("%d ", s[i][j]);
puts("");
}
return 0;
}
标签：变换 题目 solution class clu put 相同 模板 long
原文地址：https://www.cnblogs.com/Memory-of-winter/p/10224228.html
