生成排列和组合

1 . 无重复元素的全排列

当然stl有next_permutation()函数，用起来更方便

#include<cstdio>
#include<string>
using namespace std;
int n,a[100],count;
void permutation(int k){
	if(k==n){
		for(int i=1;i<=n;i++)
			printf("%d ",a[i]);
		printf("\n");
		count++;
		return;
	}
	for(int i=k;i<=n;i++){
		int tem=a[k];
		a[k]=a[i];
		a[i]=tem;
		permutation(k+1);
		tem=a[k];
		a[k]=a[i];
		a[i]=tem;
	}
}
int main(){
	while(scanf("%d",&n)!=EOF){
		for(int i=1;i<=n;i++)
			a[i]=i;
			//scanf("%d",&a[i]);
		count=0;
		permutation(1);
		printf("There is %d permutatiomns\n",count);
	}
}

2 . 有重复元素的全排列

 

#include<cstdio>
#include<string>
#include<map>
using namespace std;
int n,m,seq[100],sum[100],count,out[100];          //n表示输入字符数，m表示不重复字符数
void permutation(int k){
	if(k==n+1){
		for(int i=1;i<=n;i++)
			printf("%d ",out[i]);
		printf("\n");
		count++;
		return;
	}
	for(int i=1;i<=m;i++)
		if(sum[i]){
			sum[i]--;
			out[k]=seq[i];
			permutation(k+1);
			sum[i]++;
		}
}
int main(){
	int tem;
	while(scanf("%d",&n)!=EOF){
		map<int,int>ma;
		m=0;
		memset(sum,0,sizeof(sum));
		for(int i=1;i<=n;i++){
			scanf("%d",&tem);
			if(ma[tem]!=0)
				sum[ma[tem]]++;
			else{
				seq[++m]=tem;
				sum[m]=1;
				ma[tem]=m;
			}
		}
		count=0;
		permutation(1);
		printf("There is %d permutatiomns\n",count);
	}
}

 3 . 有重复元素的组合

#include<cstdio>
#include<string>
#include<map>
using namespace std;
int n,k,m,seq[100],sum[100],count,out[100];          //n表示输入字符数，m表示不重复字符数
void combination(int now,int p){
	if(now==k+1){
		for(int i=1;i<=k;i++)
			printf("%d ",out[i]);
		printf("\n");
		count++;
		return;
	}
	for(int i=p;i<=m;i++)
		if(sum[i]){
			sum[i]--;
			out[now]=seq[i];
			combination(now+1,i);
			sum[i]++;
		}
}
int main(){
	int tem;
	while(scanf("%d %d",&n,&k)!=EOF){
		map<int,int>ma;
		m=0;
		memset(sum,0,sizeof(sum));
		for(int i=1;i<=n;i++){
			scanf("%d",&tem);
			if(ma[tem]!=0)
				sum[ma[tem]]++;
			else{
				seq[++m]=tem;
				sum[m]=1;
				ma[tem]=m;
			}
		}
		count=0;
		combination(1,1);
		printf("There is %d combinations\n",count);
	}
}

 4 . 有重复元素的全组合

#include<cstdio>
#include<string>
#include<map>
using namespace std;
int n,k,m,seq[100],sum[100],count,out[100];          //n表示输入字符数，m表示不重复字符数
void combination(int now,int p){
	if(now>1){
		for(int i=1;i<now;i++)
			printf("%d ",out[i]);
		printf("\n");
		count++;
	}
	for(int i=p;i<=m;i++)
		if(sum[i]){
			sum[i]--;
			out[now]=seq[i];
			combination(now+1,i);
			sum[i]++;
		}
}
int main(){
	int tem;
	while(scanf("%d",&n)!=EOF){
		map<int,int>ma;
		m=0;
		memset(sum,0,sizeof(sum));
		for(int i=1;i<=n;i++){
			scanf("%d",&tem);
			if(ma[tem]!=0)
				sum[ma[tem]]++;
			else{
				seq[++m]=tem;
				sum[m]=1;
				ma[tem]=m;
			}
		}
		count=0;
		combination(1,1);
		printf("There is %d combinations\n",count);
	}
}