几个编程问题非递归解决

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问题一：

打印1,2---n中所有满足1<=i1<i2<--<ik<=n(k<=n)的有序数组(i1, i2,---ik)并输出，要求用非递归方法解决.

代码：

#include <stdio.h>#define N 6#define K 4void main(){int a[K];bool TF;int i, j;i=0;TF=false;while(1){if (i==0){if (TF==false){a[i]=1;}else{a[i]++;}if (a[i]>N-K+1)break;i++;TF=false;continue;}else{if (TF==false)a[i]=a[i-1]+1;else{a[i]++;if ((a[i]-a[i-1])>(N-a[i-1])-(K-i)+1){i--;TF=true;continue;}}if (i==K-1){for (j=0; j<K; j++){printf("%d ", a[j]);}printf("\n");TF=true;continue;}i++;TF=false;continue;}}}

问题二:打印所有n元排列,非递归解决

#include <stdio.h>#define N 5int first(int index, int number[]);void main(){int a[N]={0};int number[N]={0};int i, j, temp;bool TF;i=0;TF=false;while(1){if (i==0){if (a[i]!=0)number[a[i]-1]=0;a[i]++;if (a[i]>N)break;number[a[i]-1]=1;i++;TF=false;continue;}else{temp=first(a[i], number);if (temp==0){number[a[i]-1]=0;a[i]=0;i--;TF=true;continue;}if (TF){number[a[i]-1]=0;a[i]=temp;number[a[i]-1]=1;}else{a[i]=temp;number[a[i]-1]=1;}if (i==N-1){for (j=0; j<N; j++){printf("%d ", a[j]);}printf("\n");continue;}i++;TF=false;continue;}}}int first(int index, int number[]){int i;for (i=index; i<N; i++){if (number[i]==0)return i+1;}return 0;}

问题三：求不定方程x1+---xk=J的非负整数解

#include <stdio.h>#define K 3#define J 5void main(){int a[K]={0};bool TF;int i, j, t;TF=false;t=0;i=0;while(1){if (i==0){if (TF){a[i]++;if (a[i]>J)break;}t+=a[i];i++;TF=false;continue;}else{if (TF){a[i]++;if (a[i]>J-t){a[i]=0;t-=a[i-1];i--;TF=true;continue;}}            if (i==K-1){if (t+a[i]==J){for (j=0; j<K; j++){    printf("%d ", a[j]);}printf("\n");}TF=true;continue;}t+=a[i];i++;TF=false;continue;}}}

变体一：
求不定方程x1+---xk=J的正整数解

#include <stdio.h>#define K 2#define J 4void main(){int a[K]={0};int i, j, t;t=0;i=0;while(1){if (i==0){    a[i]++;if (a[i]>=J)break;t+=a[i];i++;continue;}else{a[i]++;if (a[i]>J-t){a[i]=0;t-=a[i-1];i--;continue;}            if (i==K-1){if (t+a[i]==J){for (j=0; j<K; j++){    printf("%d ", a[j]);}printf("\n");}continue;}t+=a[i];i++;continue;}}}

变体二：求不定方程x1+---xk=J在约束条件0<=x1<=m1----0<=xk<=mk下的非负整数解

#include <stdio.h>#define K 2#define J 5void main(){int a[K]={0};int m[K]={2, 3};bool TF;int i, j, t;TF=false;t=0;i=0;while(1){if (i==0){if (TF){a[i]++;if (a[i]>m[i] || a[i]>J)break;}t+=a[i];i++;TF=false;continue;}else{if (TF){a[i]++;if (a[i]>J-t || a[i]>m[i]){a[i]=0;t-=a[i-1];i--;TF=true;continue;}}            if (i==K-1){if (t+a[i]==J){for (j=0; j<K; j++){    printf("%d ", a[j]);}printf("\n");}TF=true;continue;}t+=a[i];i++;TF=false;continue;}}}

变体三:求不定方程x1+---xk=J在约束条件1<=x1<=m1----1<=xk<=mk下的正整数解

#include <stdio.h>#define K 2#define J 4void main(){int a[K]={0};int m[K]={2, 3};int i, j, t;t=0;i=0;while(1){if (i==0){    a[i]++;if (a[i]>m[i] || a[i]>=J)break;t+=a[i];i++;continue;}else{a[i]++;if (a[i]>J-t || a[i]>m[i]){a[i]=0;t-=a[i-1];i--;continue;}            if (i==K-1){if (t+a[i]==J){for (j=0; j<K; j++){    printf("%d ", a[j]);}printf("\n");}continue;}t+=a[i];i++;continue;}}}

附问题一的递归代码：

#include <stdio.h>#include <malloc.h>#include <stdlib.h>void DY(int *p, int k, int n);int C(int n, int k);void L(int **p1, int *p2, int k, int n);void main (){int a[5]={1, 2, 3, 4, 5};DY(a, 4 , 5);}void DY(int *p, int k, int n){int **p1;int i, j, m;j=C(n, k);p1=(int **) malloc(j*sizeof(int *));for (i=0; i<j; i++)p1[i]=(int *) malloc(k*sizeof(int));L(p1, p, k, n);for (i=0; i<j; i++){for (m=0; m<k; m++)printf ("%d", *(p1[i]+m));printf ("\n");}}int C(int n, int k){if (k==0 || k==n)return (1);return (C(n-1, k)+C(n-1, k-1));}void L(int **p1, int *p2, int k, int n){int i;if (k==1){for (i=0; i<n; i++)*p1[i]=*(p2+i);}else{if (k==n){for (i=0; i<n; i++)*(p1[0]+i)=*(p2+i);}else{int j, c, d, z1, z2;int **p;j=0; d=1; c=1;for (i=n-k; i>=0; i--){p=(int **) malloc(c*sizeof(int *));for (z1=0; z1<c; z1++)       p[z1]=(int *) malloc((k-1)*sizeof(int));                L(p, p2+i+1, k-1, n-i-1);for (z1=0; z1<c; z1++)for (z2=0; z2<(k-1); z2++)*(p1[z1+j]+z2+1)=*(p[z1]+z2);while (j<d) {*p1[j]=*(p2+i);j++;}if (i!=0){    c=(c*(n-i))/(n-i-k+1);      d=d+c;}}}}}

问题二的递归解法就是一道常规习题

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