排列组合（三）

1 . 无重复元素的全排列

当然stl有next_permutation()函数，用起来更方便

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1. #include<cstdio>  
2. #include<string>  
3. using namespace std;  
4. int n,a[100],count;  
5. void permutation(int k){  
6.     if(k==n){  
7.         for(int i=1;i<=n;i++)  
8.             printf("%d ",a[i]);  
9.         printf("\n");  
10.         count++;  
11.         return;  
12.     }  
13.     for(int i=k;i<=n;i++){  
14.         int tem=a[k];  
15.         a[k]=a[i];  
16.         a[i]=tem;  
17.         permutation(k+1);  
18.         tem=a[k];  
19.         a[k]=a[i];  
20.         a[i]=tem;  
21.     }  
22. }  
23. int main(){  
24.     while(scanf("%d",&n)!=EOF){  
25.         for(int i=1;i<=n;i++)  
26.             a[i]=i;  
27.             //scanf("%d",&a[i]);  
28.         count=0;  
29.         permutation(1);  
30.         printf("There is %d permutatiomns\n",count);  
31.     }  
32. }  

#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

 

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1. #include<cstdio>  
2. #include<string>  
3. #include<map>  
4. using namespace std;  
5. 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  
6. void permutation(int k){  
7.     if(k==n+1){  
8.         for(int i=1;i<=n;i++)  
9.             printf("%d ",out[i]);  
10.         printf("\n");  
11.         count++;  
12.         return;  
13.     }  
14.     for(int i=1;i<=m;i++)  
15.         if(sum[i]){  
16.             sum[i]--;  
17.             out[k]=seq[i];  
18.             permutation(k+1);  
19.             sum[i]++;  
20.         }  
21. }  
22. int main(){  
23.     int tem;  
24.     while(scanf("%d",&n)!=EOF){  
25.         map<int,int>ma;  
26.         m=0;  
27.         memset(sum,0,sizeof(sum));  
28.         for(int i=1;i<=n;i++){  
29.             scanf("%d",&tem);  
30.             if(ma[tem]!=0)  
31.                 sum[ma[tem]]++;  
32.             else{  
33.                 seq[++m]=tem;  
34.                 sum[m]=1;  
35.                 ma[tem]=m;  
36.             }  
37.         }  
38.         count=0;  
39.         permutation(1);  
40.         printf("There is %d permutatiomns\n",count);  
41.     }  
42. }  

#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

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1. #include<cstdio>  
2. #include<string>  
3. #include<map>  
4. using namespace std;  
5. 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  
6. void combination(int now,int p){  
7.     if(now==k+1){  
8.         for(int i=1;i<=k;i++)  
9.             printf("%d ",out[i]);  
10.         printf("\n");  
11.         count++;  
12.         return;  
13.     }  
14.     for(int i=p;i<=m;i++)  
15.         if(sum[i]){  
16.             sum[i]--;  
17.             out[now]=seq[i];  
18.             combination(now+1,i);  
19.             sum[i]++;  
20.         }  
21. }  
22. int main(){  
23.     int tem;  
24.     while(scanf("%d %d",&n,&k)!=EOF){  
25.         map<int,int>ma;  
26.         m=0;  
27.         memset(sum,0,sizeof(sum));  
28.         for(int i=1;i<=n;i++){  
29.             scanf("%d",&tem);  
30.             if(ma[tem]!=0)  
31.                 sum[ma[tem]]++;  
32.             else{  
33.                 seq[++m]=tem;  
34.                 sum[m]=1;  
35.                 ma[tem]=m;  
36.             }  
37.         }  
38.         count=0;  
39.         combination(1,1);  
40.         printf("There is %d combinations\n",count);  
41.     }  
42. }  

#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

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1. #include<cstdio>  
2. #include<string>  
3. #include<map>  
4. using namespace std;  
5. 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  
6. void combination(int now,int p){  
7.     if(now>1){  
8.         for(int i=1;i<now;i++)  
9.             printf("%d ",out[i]);  
10.         printf("\n");  
11.         count++;  
12.     }  
13.     for(int i=p;i<=m;i++)  
14.         if(sum[i]){  
15.             sum[i]--;  
16.             out[now]=seq[i];  
17.             combination(now+1,i);  
18.             sum[i]++;  
19.         }  
20. }  
21. int main(){  
22.     int tem;  
23.     while(scanf("%d",&n)!=EOF){  
24.         map<int,int>ma;  
25.         m=0;  
26.         memset(sum,0,sizeof(sum));  
27.         for(int i=1;i<=n;i++){  
28.             scanf("%d",&tem);  
29.             if(ma[tem]!=0)  
30.                 sum[ma[tem]]++;  
31.             else{  
32.                 seq[++m]=tem;  
33.                 sum[m]=1;  
34.                 ma[tem]=m;  
35.             }  
36.         }  
37.         count=0;  
38.         combination(1,1);  
39.         printf("There is %d combinations\n",count);  
40.     }  
41. }  

#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);}}