HDU 4965 Fast Matrix Calculation 矩阵乘法 乘法结合律

一种奇葩的写法，纪念一下当时的RE。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <cmath>
  6 #include <algorithm>
  7 #include <string>
  8 #include <queue>
  9 #include <stack>
 10 #include <vector>
 11 #include <map>
 12 #include <set>
 13 #include <functional>
 14 #include <cctype>
 15 #include <time.h>
 16
 17 using namespace std;
 18
 19 const int INF = 1<<30;
 20 const int MAXN = 1005;
 21
 22 struct Matrix {
 23     int a[MAXN][MAXN];
 24     int col, row;
 25 };
 26
 27 Matrix tMu;
 28
 29 void multiplication(const Matrix &x, const Matrix &y) {
 30     tMu.row= x.row; tMu.col = y.col;
 31     for (int i = 0; i < tMu.row; i++)
 32         for (int j = 0; j < tMu.row; j++) {
 33             tMu.a[i][j] = 0;
 34             for (int k = 0; k < x.col; k++)
 35                 tMu.a[i][j] = (tMu.a[i][j]+x.a[i][k]*y.a[k][j])%6;
 36         }
 37 }
 38
 39 Matrix a, b, ONE;
 40 Matrix t;
 41 Matrix tmp, res;
 42
 43 int n, k;
 44
 45 void pow(const Matrix &x, int d) {
 46     res = ONE; res.col = res.row = x.col;
 47     tmp = x;
 48     while (d>0) {
 49         if (d&1) {
 50             multiplication(res, tmp);
 51             res = tMu;
 52         }
 53         multiplication(tmp, tmp);
 54         tmp = tMu;
 55         d >>= 1;
 56     }
 57 }
 58
 59 void solve() {
 60     multiplication(b, a);
 61     t = tMu;
 62     pow(t, n*n-1);
 63     t = res;
 64     multiplication(a, t);
 65     t = tMu;
 66     multiplication(t, b);
 67     t = tMu;
 68     int ans = 0;
 69     for (int i = 0; i < n; i++) {
 70         for (int j = 0; j < n; j++)
 71             ans += t.a[i][j];
 72     }
 73     printf("%d\n", ans);
 74 }
 75
 76 int main() {
 77     #ifdef Phantom01
 78         freopen("HDU4965.txt", "r", stdin);
 79     #endif //Phantom01
 80
 81     for (int i = 0; i < MAXN; i++) {
 82         for (int j = 0; j < MAXN; j++)
 83             ONE.a[i][j] = 0;
 84         ONE.a[i][i] = 1;
 85     }
 86
 87     while (scanf("%d%d", &n, &k)!=EOF && !(n==0&&k==0)) {
 88         a.row = b.col = n;
 89         a.col = b.row = k;
 90         for (int i = 0; i < n; i++)
 91             for (int j = 0; j < k; j++)
 92                 scanf("%d", &a.a[i][j]);
 93         for (int i = 0; i < k; i++)
 94             for (int j = 0; j < n; j++)
 95                 scanf("%d", &b.a[i][j]);
 96         solve();
 97     }
 98
 99     return 0;
100 }