hdu 3037 Saving Beans lucas定理

题目链接

给n, m, p， 求sigma(i = 0 to m) C(n-1+i, n-1)%p的值。

C(n-1, n-1)+C(n-1+1, n-1)+C(n-1+2, n-1)+.......

= C(n-1, 0)+C(n, 1)+C(n+1, 2)+....

= C(n, 0)+C(n, 1)+C(n+1, 2)+....

根据C(n, k) = C(n-1, k-1)+C(n-1, k) 可以推出来上面的式子最终合并为C(n+m, m)%p。

这个式子直接用lucas定理就可以了。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
ll mod;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
ll pow(ll a, ll b) {
ll ret = 1;
while(b) {
if(b&1) {
ret = ret*a%mod;
}
a = a*a%mod;
b >>= 1;
}
return ret;
}
ll C(ll n, ll k) {
if(n<k)
return 0;
if(k>n-k)
k = n-k;
ll s1 = 1, s2 = 1;
for(int i = 0; i<k; i++) {
s1 = s1*(n-i)%mod;
s2 = s2*(i+1)%mod;
}
return s1*pow(s2, mod-2)%mod;
}
ll lucas(ll n, ll k) {
if(k == 0)
return 1;
return lucas(n/mod, k/mod)*C(n%mod, k%mod)%mod;
}
int main()
{
int t;
ll n, m;
cin>>t;
while(t--) {
scanf("%I64d%I64d%I64d", &n, &m, &mod);
ll ans = lucas(n+m, m);
printf("%I64d\n", ans);
}
return 0;
}