a,b较小时 2000
#include<iostream>
using namespace std;
const int N = 2010, mod = 1e9 + 7;
int c[N][N];
int n;
int main()
{
cin >> n;
for(int i = 0; i < N; i++)
{
for(int j = 0; j <= i; j++)
{
if(j == 0) c[i][j] = 1;
else if(j == i) c[i][j] = 1;
else c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
}
}
while(n--)
{
int x, y;
cin >> x >> y;
cout << c[x][y] << endl;
}
return 0;
}
a,b较大时 利用乘法逆元求 a,b为10^5
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
const int N = 100010, mod = 1e9 + 7;
int n;
int fact[N], infact[N];
int quick_mul(int a, int b, int mod)
{
int ans = 1 % mod;
while (b)
{
if (b & 1)
ans = (ll)ans * a % mod;
a = (ll)a * a % mod;
b >>= 1;
}
return ans;
}
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
cin >> n;
fact[0] = infact[0] = 1;
for (int i = 1; i < N; i++)
{
fact[i] = (ll)fact[i - 1] * i % mod;
infact[i] = (ll)infact[i - 1] * quick_mul(i, mod - 2, mod) % mod;
}
while (n--)
{
int a, b;
cin >> a >> b;
cout << 1ll * fact[a] * infact[b] % mod * infact[a - b] % mod << endl;
}
return 0;
}
a, b 10^18, 利用卢卡斯定理
原始定义,实现C函数
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int n;
ll a, b, p;
int quick_mul(ll a, ll b, ll mod)
{
ll ans = 1 % mod;
while (b)
{
if (b & 1)
ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
ll C(ll a, ll b, ll p)
{
if (a < b)
return 0;
ll ans = 1;
for (ll i = 1, j = a; i <= b; i++, j--)
{
ans = ans * j % p;
ans = ans * quick_mul(i, p - 2, p) % p;
}
return ans;
}
ll lucas(ll a, ll b, ll p)
{
if (a < b)
return 0;
if (a < p && b < p)
return C(a, b, p);
return C(a % p, b % p, p) * lucas(a / p, b / p, p) % p;
}
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
cin >> n;
while (n--)
{
cin >> a >> b >> p;
cout << lucas(a, b, p) << endl;
}
return 0;
}
当不用取模时,用阶乘分解加大数据乘法
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 5010;
int n, a, b;
int sum[N];
int prime[N], cnt;
bool vis[N];
void get_primes(int n)
{
for (int i = 2; i <= n; i++)
{
if (!vis[i])
prime[cnt++] = i;
for (int j = 0; prime[j] <= n / i; j++)
{
vis[prime[j] * i] = true;
if (i % prime[j] == 0)
break;
}
}
}
int get_num(int n, int p)
{
int sum = 0;
while (n)
{
sum += n / p;
n /= p;
}
return sum;
}
vector<int> mul(vector<int> a, int b)
{
int t = 0;
vector<int> ans;
for (int i = 0; i < a.size(); i++)
{
t += a[i] * b;
ans.push_back(t % 10);
t /= 10;
}
while (t)
{
ans.push_back(t % 10);
t /= 10;
}
return ans;
}
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
cin >> a >> b;
get_primes(a);
for (int i = 0; i < cnt; i++)
{
int p = prime[i];
sum[i] = get_num(a, p) - get_num(b, p) - get_num(a - b, p);
}
vector<int> ans;
ans.push_back(1);
for (int i = 0; i < cnt; i++)
for (int j = 0; j < sum[i]; j++)
ans = mul(ans, prime[i]);
for (int i = ans.size() - 1; i >= 0; i--)
cout << ans[i];
cout << endl;
return 0;
}