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UVaLive 7455 Linear Ecosystem (Gaussi 消元)
题意：对一个k元向量, 每次左乘一个k*k的矩阵得到新的向量.问经过一定次数的左乘后，能否使得该向量不再变化. (同时要求此时向量非零)。
析：设初始向量为A，矩阵为P.由于每次矩阵P都是左乘A, 那么可以把若干个P合并. 则题目的条件是：
化简为： 
由于要求 
所以 P-1 必须不可逆.可以直接用高斯消元求P-1的秩，判断是否可逆（满秩即可逆）.
所以这个题，并不用求解，只要判断秩就好。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e2 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}double a[25][25];bool Gauss(){ int ans = 0, r = 0; for(int i = 0; i < n; ++i){ for(int j = r; j < n; ++j) if(fabs(a[j][i]) > eps){ for(int k = i; k < n; ++k) swap(a[j][k], a[r][k]); break; } if(fabs(a[r][i]) < eps){ ++ans; continue; } for(int j = 0; j < n; ++j) if(j != r && fabs(a[j][i]) > eps){ double tmp = a[j][i]/a[r][i]; for(int k = i; k < n; ++k) a[j][k] -= tmp * a[r][k]; } ++r; } return ans;}int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ scanf("%d", &n); for(int i = 0; i < n; ++i){ for(int j = 0; j < n; ++j) scanf("%lf", a[i]+j); a[i][i] -= 1.0; } printf("%d", Gauss()); if(kase == T) continue; printf("%c", kase % 5 ? ‘ ‘ : ‘\n‘); } if(T % 5) printf("\n"); return 0;}
UVaLive 7455 Linear Ecosystem (Gaussi 消元)
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