标签：dia return gre font number output efi 表示 n个元素
从n个元素里取出m个，记C(n,m); 原始公式：C(n,m) = n! / ( m!*(n-m)! )
在编程里，初中高中的阶乘公式已经不太管用了,数据一下子就爆。
公式1：C(n,m) = (n-m+1) * C(n,m-1) / m
用数组存放，从C(n,0)开始存到C(n,m),数组下标表示m的值
且看题目：（裸题）
In how many ways can you choose k elements out of n elements, not taking order into account?
Input
Output
Sample Input
4 2 10 5 49 6 0 0
Sample Output
6 252 13983816
代码：
#include<iostream> #include<stdio.h> #include<algorithm> using namespace std; long long a[9999999]; int main()///裸题，单纯套公式 { long long n,k,i; while(cin>>n>>k&&(n+k)) { a[0]=1; if(k>n-k) k=n-k; for(i=1;i<=k;i++) a[i]=(n-i+1)*a[i-1]/i; cout<<a[k]<<endl; } return 0; }
公式2：C(n,m) = C(n-1,m) + C(n-1,m-1)
计算较大的组合数，用二维数组存放各个组合数的大小
且看裸题：
Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
Input
Output
Sample Input
100 6 20 5 18 6 0 0
Sample Output
100 things taken 6 at a time is 1192052400 exactly. 20 things taken 5 at a time is 15504 exactly. 18 things taken 6 at a time is 18564 exactly.
代码：
#include<iostream> #include<stdio.h> #include<algorithm> using namespace std; long long c[101][101]; int main() { for(int i=1;i<101;i++) { c[i][0]=c[0][i]=1; c[i][i]=1; //不用担心下面n和k相等没法进入循环的情况 } for(int n=1;n<101;n++) for(int k=1;k<n;k++)//一直是k<=n；不过等于的情况解决了 c[n][k]=c[n-1][k]+c[n-1][k-1]; int n,k; while(cin>>n>>k&&(n+k)) { printf("%d things taken %d at a time is %d exactly.\n",n,k,c[n][k]); } return 0; }
数据更大，需要求模，一般用鲁卡斯定理：
C(n,m)%p p为素数，总有：
m = t*p + r
n = s*p + q → C(n,m)%p = C(s,t)%p * C(q,r)%p → C(n,m)%p = C(n/p,m/p)%p * C(m%p,n%p)%p
题目：
InputThe first line contains one integer T, means the number of cases.
Then followed T lines, each line contains three integers n, m, p, means that squirrels will save no more than m same
beans in n different trees, 1 <= n, m <= 1000000000, 1 < p < 100000 and p is guaranteed to be a prime.OutputYou
should output the answer modulo p.
Sample Input
2
1 2 5 2 1 5
Sample Output
3 3
Hint For sample 1, squirrels will put no more than 2 beans in one tree. Since trees are different, we can label them as 1, 2 … and so on. The 3 ways are: put no beans, put 1 bean in tree 1 and put 2 beans in tree 1. For sample 2, the 3 ways are: put no beans, put 1 bean in tree 1 and put 1 bean in tree 2.
思路：由于树上可以不放豆，C(n+m,m)。数据太大，需要很多防爆措施，快速幂，逆元，阶乘，鲁卡斯定理。
代码：
#include<iostream> #include<stdio.h> #include<algorithm> #define ll long long using namespace std; ll n, m, p; ll fac[1000000+10]={1,1}; ///0和1的阶乘为1 void getfac() ///打印阶乘表 { for(ll i=1;i<=p;i++) ///每次打到p就好 fac[i]=fac[i-1]*i%p; } ll power(ll a,ll b,ll p) ///大数据逆元计算又需要快速幂 { ///a的b次方对p求模 ll ans=1; while(b>0) { if(b%2==1) ans=ans*a%p; b>>=1; a=a*a%p; } return ans%p; } ll C(ll n, ll m ,ll p) { ///C(n,m) = n! / ( m!*(n-m)! ) ///数据太大肯定爆，p又是素数。换成求m!*(n-m)!的逆元，又不能一起求，会爆数据，分开求，看做n!/m! * 1/(n-m)! ///由于是对p求模，n，m范围在阶乘表范围里 if(m>n) return 0; return fac[n]%p * power(fac[m], p-2, p) * power(fac[n-m], p-2, p) % p; } ll lucase(ll n, ll m ) { if(m==0) return 1; else return C(n%p, m%p, p)* lucase(n/p,m/p)%p;/// (n/p),(m/p)不一定比p小，继续用鲁卡斯 } int main() { int t; cin>>t; while(t--) { cin>>n>>m>>p; getfac(); cout<<lucase(n+m,n)<<endl; } return 0; }
标签：dia return gre font number output efi 表示 n个元素
原文地址：https://www.cnblogs.com/shoulinniao/p/9426905.html
