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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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int n, m, k;
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int a[N][N], b[N][N], c[N][N];
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// 矩阵乘法模板
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void mul(int a[][N], int b[][N], int c[][N]) {
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// 注意:这里的临时数组t绝不是画蛇添足!
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// 因为调用的时候,有时会传递mul(a,b,b)这样的东东,如果每次memset(c,0,sizeof c),
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// 就会造成b在运行前被清空,导致结果错误
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// 代码解释:
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// 从结果出发理解:
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// C(i,j) =A(i,1)×B(1,j)+A(i,2)×B(2,j) +...A(r,1)×B(r,j)
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// 抽象一下:
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// C(i,j) = C(i,j) + A(i,k)× B(k,j)
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int t[N][N] = {0};
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for (int i = 0; i < N; i++)
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for (int j = 0; j < N; j++)
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for (int k = 0; k < N; k++)
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t[i][j] = t[i][j] + a[i][k] * b[k][j]; // 矩阵乘法
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memcpy(c, t, sizeof t);
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("B2105.in", "r", stdin);
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#endif
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cin >> n >> m >> k;
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// A矩阵 n*m
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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cin >> a[i][j];
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// B矩阵m*k
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for (int i = 1; i <= m; i++)
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for (int j = 1; j <= k; j++)
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cin >> b[i][j];
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// 矩阵乘法
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mul(a, b, c);
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// 输出结果,控制格式
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for (int i = 1; i <= n; i++) {
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for (int j = 1; j <= k; j++)
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printf("%d ", c[i][j]);
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printf("\n");
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}
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return 0;
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}
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