标签:
组合数专题
直接贴上代码,以后再补充。。。。。。
方法1:
const int N = 1e5 + 10;
const int MOD = 1e9 + 7;
int f[N], finv[N], inv[N];
void init(void) {
inv[1] = 1;
for (int i=2; i<N; ++i) {
inv[i] = (MOD - MOD / i) * 1ll * inv[MOD%i] % MOD;
}
f[0] = finv[0] = 1;
for (int i=1; i<N; ++i) {
f[i] = f[i-1] * 1ll * i % MOD;
finv[i] = finv[i-1] * 1ll * inv[i] % MOD;
}
}
int comb(int n, int k) { //C (n, k) % MOD
if (k < 0 || k > n) return 0;
return f[n] * 1ll * finv[n-k] % MOD * finv[k] % MOD;
}
方法2:
const int N = 2000 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int comb[N][N];
void init(void) {
for (int i=0; i<N; ++i) {
comb[i][i] = comb[i][0] = 1;
for (int j=1; j<i; ++j) {
comb[i][j] = comb[i-1][j] + comb[i-1][j-1];
if (comb[i][j] >= MOD) {
comb[i][j] -= MOD;
}
}
}
//printf ("%d\n", comb[6][3]);
}
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原文地址:http://www.cnblogs.com/Running-Time/p/4749067.html