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59 lines
1.5 KiB
59 lines
1.5 KiB
2 years ago
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#include <set>
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#include <map>
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#include <cmath>
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#include <stack>
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#include <queue>
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#include <vector>
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#include <string>
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#include <cstdio>
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#include <cstring>
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#include <sstream>
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#include <iomanip>
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#include <iostream>
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#include <algorithm>
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#define clr(str, x) memset(str, x, sizeof(str))
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#define FRER() freopen("in.txt", "r", stdin);
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#define FREW() freopen("out.txt", "w", stdout);
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#define MAX_INF 0x7fffffff
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#define INF 0x3f3f3f3f
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#define maxn
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typedef long long int ll;
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using namespace std;
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const int mod = 1000007;
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int C[405][405];
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int n, m, k, ans = 0;
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void init() {
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for (int i = 1; i <= 400; i++) {
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C[i][0] = C[i][i] = 1;
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for (int j = 1; j < i; j++)
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C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
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}
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}
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int main() {
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// FRER()
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// FREW()
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init();
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int T, kase = 1;
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int s1, s2, s3, s4, s5;
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scanf("%d", &T);
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while (T--) {
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ans = 0;
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scanf("%d%d%d", &n, &m, &k);
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if (k == 0) {
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printf("Case %d: 0\n", kase++);
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continue;
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}
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s1 = C[n * m][k];
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s2 = 2 * (C[n * (m - 1)][k] + C[(n - 1) * m][k]) % mod;
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s3 = (C[(n - 2) * m][k] + 4 * C[n * m - n - m + 1][k] + C[n * (m - 2)][k]) % mod;
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s4 = 2 * (C[(n - 2) * (m - 1)][k] + C[(n - 1) * (m - 2)][k]) % mod;
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s5 = C[(n - 2) * (m - 2)][k] % mod;
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ans = s1 - s2 + s3 - s4 + s5;
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ans = (ans + 10 * mod) % mod; // 防止取余后出现负数
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printf("Case %d: %d\n", kase++, ans);
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}
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return 0;
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}
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