求组合个数 POJ2249 以及一种组合的输出方法

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#integer #input #output #less #each #优化

本文介绍了一种计算从N个不同元素中选取M个元素的组合数的方法，并提供两种实现方案：一种是直接计算分子分母后相除，另一种通过约分减少计算复杂度。此外，还提供了一个输出所有可能组合的示例程序。

Description

In how many ways can you choose k elements out of n elements, not taking order into account?
Write a program to compute this number.

Input

The input will contain one or more test cases.
Each test case consists of one line containing two integers n (n>=1) and k (0<=k<=n).
Input is terminated by two zeroes for n and k.

Output

For each test case, print one line containing the required number. This number will always fit into an integer, i.e. it will be less than 2³¹.
Warning: Don't underestimate the problem. The result will fit into an integer - but if all intermediate results arising during the computation will also fit into an integer depends on your algorithm. The test cases will go to the limit.

Sample Input

Sample Output

6
252
13983816

此题两种解决方法，第一种为普通方法，第二种则为约分方式

普通方法

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>

using namespace std;

int main()
{
	long long a,b,i,sum1,sum2;
	//freopen("E://input.txt","r",stdin);
	scanf("%lld%lld",&a,&b);
	while(!(a==0&&b==0))
	{
		sum1=sum2=1;
		if(b>a-b)
            b=a-b;  //注意本行优化，时间TLE直降0ms
		for (i=a;i>a-b;i--)
		{
			sum1*=i;
		    sum1/=sum2;
			sum2++;
		}
		printf("%lld/n",sum1);
		scanf("%lld%lld",&a,&b);
	}
	return 0;
}

约分法

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>

using namespace std;

int gcd(int a,int b) {
    if(b==0) return(a);
    return(gcd(b,a%b));
}

int main()
{
	//freopen("E://input.txt","r",stdin);
    int i,j,g,k,n,u[40];
    while(scanf("%d%d",&n,&k),n) {
        if(k>(n/2)) k=n-k;
        for(i=0;i<k;i++) u[i]=n+i+1-k;
        for(j=2;j<=k;j++) {
            n=j;
            for(i=0;(i<k)&&(n>1);i++) {
                g=gcd(u[i],n);
                u[i]/=g;
                n/=g;
            }
        }
        n=1;
        for(i=0;i<k;i++) n*=u[i];
        printf("%d/n",n);
    }
	return 0;
}

输出N个数中取M的各种组合方式

#include <stdio.h> #include <stdlib.h> #include <iostream> #include <string.h> using namespace std; int num,ResMatrix[15]; void GetRes(int s,int d) { int i,j; for (i=s;i>=d;i--) { ResMatrix[d]=i; if (d==1) { for (j=num;j>=1;j--) printf("%d",ResMatrix[j]); printf("/n"); } else GetRes(i-1,d-1); } } int main() { int n; scanf("%d%d",&n,&num); GetRes(n,num); return 0; }