算法1
适用数据范围
1≤n≤10000,
1≤b≤a≤2000
核心代码
c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod
原理
类似于动态规划(dp),将情况划分为选第j个和不选第j个
选第j个可以看作从剩下的i-1里再选j-1个 即c[i - 1][j - 1]
不选第j个可以看作从剩下的i-1里再选个 即c[i - 1][j]
C++ 代码
注意初始化外层循环i要从0开始
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 2010;
const int mod = 1e9 + 7;
int n;
int c[N][N];
void init(){
for(int i = 0; i <= 2000; i ++){
for(int j = 0; j <= i; j ++){
if(j == 0) c[i][j] = 1;
else c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
}
}
}
int main(){
init();
cin >> n;
while(n --){
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",c[a][b]);
}
return 0;
}
算法2
适用数据范围
1≤n≤10000,
1≤b≤a≤105
原理
C++ 代码
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
const int N = 100010, mod = 1e9 + 7;
int fact[N],infact[N];
int qmi(int a,int b,int p){ //快速幂
int res = 1;
while(b){
if(b & 1) res = (LL)res * a % p;
b >>= 1;
a = (LL)a * a % p;
}
return res;
}
int main(){
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] * qmi(i,mod - 2,mod) % mod;
}
int n;
scanf("%d",&n);
while(n --){
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",(LL)fact[a] * infact[a - b] % mod * infact[b] % mod);
}
return 0;
}
算法3
适用数据范围
1≤n≤20,
1≤b≤a≤1018,
1≤p≤105
原理
卢卡斯定理式子
C++ 代码
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 100010;
typedef long long LL;
int p;
int fact[N],infact[N];
int qmi(int a,int b,int p){
int res = 1;
while(b){
if(b & 1) res = (LL)res * a % p;
b >>= 1;
a = (LL)a * a % p;
}
return res;
}
int C(int a, int b, int p)
{
if (b > a) return 0;
int res = 1;
for (int i = 1, j = a; i <= b; i ++, j -- )
{
res = (LL)res * j % p;
res = (LL)res * qmi(i, p - 2, p) % p;
}
return res;
}
int lucas(LL a,LL b,int p){
if(a < p && b < p) return C(a,b,p);
return (LL)C(a % p,b % p,p) * lucas(a / p,b / p,p) % p;
}
int main(){
int n;
cin >> n;
while(n --){
LL a,b;
cin >> a >> b >> p;
cout << lucas(a,b,p) << endl;
}
return 0;
}
算法4
适用于不取模 直接求某个组合数的值
原理
大整数乘法 大整数除法
C++代码
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 5010;
int primes[N], cnt;
int sum[N];
bool st[N];
void get_primes(int n)
{
for (int i = 2; i <= n; i ++ )
{
if (!st[i]) primes[cnt ++ ] = i;
for (int j = 0; primes[j] <= n / i; j ++ )
{
st[primes[j] * i] = true;
if (i % primes[j] == 0) break;
}
}
}
int get(int n, int p)
{
int res = 0;
while (n)
{
res += n / p;
n /= p;
}
return res;
}
vector<int> mul(vector<int> a, int b)
{
vector<int> c;
int t = 0;
for (int i = 0; i < a.size(); i ++ )
{
t += a[i] * b;
c.push_back(t % 10);
t /= 10;
}
while (t)
{
c.push_back(t % 10);
t /= 10;
}
return c;
}
int main()
{
int a, b;
cin >> a >> b;
get_primes(a);
for (int i = 0; i < cnt; i ++ )
{
int p = primes[i];
sum[i] = get(a, p) - get(a - b, p) - get(b, p);
}
vector<int> res;
res.push_back(1);
for (int i = 0; i < cnt; i ++ )
for (int j = 0; j < sum[i]; j ++ )
res = mul(res, primes[i]);
for (int i = res.size() - 1; i >= 0; i -- ) printf("%d", res[i]);
puts("");
return 0;
}
好刘的蓝铜
hh