Monday, June 6, 2016

求组合数的算法 - Cppowboy's Blog - SegmentFault


求组合数的算法 - Cppowboy's Blog - SegmentFault

问题:求解组合数C(n,m),即从n个相同物品中取出m个的方案数,由于结果可能非常大,对结果模10007即可。

暴力求解,C(n,m)=n(n-1)…*(n-m+1)/m!
int Combination(int n, int m)
{
const int M = 10007;
int ans = 1;
for(int i=n; i>=(n-m+1); --i)
ans *= i;
while(m)
ans /= m--;
return ans % M;
}
这种方案的缺陷是,在计算过程中很快ans就溢出了,一般情况下,n不能超过12。补救办法之一是将先乘后除改为交叉地进行乘除,先除能整除的,但也只能满足n稍微增大的情况,n最多只能满足两位数。补救办法之二是换用高精度运算,这样结果不会有问题,只是需要实现大数相乘、相除和取模等运算,实现起来比较麻烦,时间复杂度为O(n)。

方案二

打表,C(n,m)=C(n-1,m-1)+C(n-1,m)
由于组合数满足以上性质,可以预先生成所有用到的组合数,使用时,直接查找即可。生成的复杂度为O(n^2),查询复杂度为O(1)。较方案一而言,支持的数量级大有提升,在1秒内,基本能处理10000以内的组合数。算法的预处理时间较长,另外空间花费较大,都是平方级的,优点是实现简单,查询时间快。
const int M = 10007;
const int MAXN = 1000;
int C[MAXN+1][MAXN+1];
void Initial()
{
int i,j;
for(i=0; i<=MAXN; ++i)
{
C[0][i] = 0;
C[i][0] = 1;
}
for(i=1; i<=MAXN; ++i)
{
for(j=1; j<=MAXN; ++j)
C[i][j] = (C[i-1][j] + C[i-1][j-1]) % M;
}
}

int Combination(int n, int m)
{
return C[n][m];
}

方案三

质因数分解,C(n,m)=n!/(m!*(n-m)!),设n!分解因式后,质因数p的次数为a;对应地m!分解后p的次数为b;(n-m)!分解后p的次数为c;则C(n,m)分解后,p的次数为a-b-c。计算出所有质因子的次数,它们的积即为答案,即C(n,m)=p1 a1-b1-c1p2 a2-b2-c2…pk ak-bk-ck。n!分解后p的次数为:n/p+n/p 2+…+n/p k。
算法的时间复杂度比前两种方案都低,基本上跟n以内的素数个数呈线性关系,而素数个数通常比n都小几个数量级,例如100万以内的素数不到8万个。用筛法生成素数的时间接近线性。该方案1秒钟能计算 1kw数量级的组合数。如果要计算更大,内存和时间消耗都比较大。
//用筛法生成素数
const int MAXN = 1000000;
bool arr[MAXN+1] = {false};
vector<int> produce_prim_number()
{
vector<int> prim;
prim.push_back(2);
int i,j;
for(i=3; i*i<=MAXN; i+=2)
{
if(!arr[i])
{
prim.push_back(i);
for(j=i*i; j<=MAXN; j+=i)
arr[j] = true;
}
}
while(i<=MAXN)
{
if(!arr[i])
prim.push_back(i);
i+=2;
}
return prim;
}

//计算n!中素因子p的指数
int Cal(int x, int p)
{
int ans = 0;
long long rec = p;
while(x>=rec)
{
ans += x/rec;
rec *= p;
}
return ans;
}

//计算n的k次方对M取模,二分法
int Pow(long long n, int k, int M)
{
long long ans = 1;
while(k)
{
if(k&1)
{
ans = (ans * n) % M;
}
n = (n * n) % M;
k >>= 1;
}
return ans;
}

//计算C(n,m)
int Combination(int n, int m)
{
        const int M = 10007;
vector<int> prim = produce_prim_number();
long long ans = 1;
int num;
for(int i=0; i<prim.size() && prim[i]<=n; ++i)
{
num = Cal(n, prim[i]) - Cal(m, prim[i]) - Cal(n-m, prim[i]);
ans = (ans * Pow(prim[i], num, M)) % M;
}
return ans;
}

方案四

Lucas定理,设p是一个素数(题目中要求取模的数也是素数),将n,m均转化为p进制数,表示如下:
满足下式:
即C(n,m)模p等于p进制数上各位的C(ni,mi)模p的乘积。利用该定理,可以将计算较大的C(n,m)转化成计算各个较小的C(ni,mi)。
该方案能支持整型范围内所有数的组合数计算,甚至支持64位整数,注意中途溢出处理。该算法的时间复杂度跟n几乎不相关了,可以认为算法复杂度在常数和对数之间。
#include <stdio.h> const int M = 10007; int ff[M+5]; //打表,记录n!,避免重复计算 //求最大公因数 int gcd(int a,int b) { if(b==0) return a; else return gcd(b,a%b); } //解线性同余方程,扩展欧几里德定理 int x,y; void Extended_gcd(int a,int b) { if(b==0) { x=1; y=0; } else { Extended_gcd(b,a%b); long t=x; x=y; y=t-(a/b)*y; } } //计算不大的C(n,m) int C(int a,int b) { if(b>a) return 0; b=(ff[a-b]*ff[b])%M; a=ff[a]; int c=gcd(a,b); a/=c; b/=c; Extended_gcd(b,M); x=(x+M)%M; x=(x*a)%M; return x; } //Lucas定理 int Combination(int n, int m) { int ans=1; int a,b; while(m||n) { a=n%M; b=m%M; n/=M; m/=M; ans=(ans*C(a,b))%M; } return ans; } int main(void) { int i,m,n; ff[0]=1; for(i=1;i<=M;i++) //预计算n! ff[i]=(ff[i-1]*i)%M; scanf("%d%d",&n, &m); printf("%d\n",func(n,m)); return 0; }
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