Recursion, enumeration

1. Hanoi Tower Problem

#include <stdio.h>voidHanint,Char,Char,Char );intMain () {intN; CharA='A', b='B', c='C'; scanf ("%d",&N); Han (n,a,b,c);}voidmovCharACharBB) {printf ("%c=>%c\n", A, b);}voidHanintNCharACharBCharc) {    if(n==1) mov (A,C); Else{han (n-1, a,c,b);        MOV (A,C); Han (n-1, B,a,c); }}

2. Conversion problems in the system

Enter two numbers, one representing the existing number of eg, and another that represents the few binary ch to convert. Use the EG mode ch and save it with an array.

#include <stdio.h>intMain () {inta[10000]; intEg,ch; inti,j; scanf ("%d%d",&eg,&ch);  for(i=0; i<10000; i++) A[i]=0;  for(i=0; i<10000; i++) {A[i]=eg%ch; Eg=eg/ch; }     for(i=9999; i>-1; i--)        if(a[i]!=0) Break;  for(J=i; j>-1; j--) printf ("%d", A[j]);}

3. Eight Queen's question

Define a a[8] array to save, determine if the condition is met, a[j]==i| available | i-t==a[j]-j| | T+i==a[j]+j

motioned as follows

0-1-2-3-4-5-6-7
1 0-1-2-3-4-5-6
2 1 0-1-2-3-4-5
3 2 1 0-1-2-3-4
4 3 2 1 0-1-2-3
5 4 3 2 1 0-1-2
6 5 4 3 2 1 0-1
7 6 5 4 3 2 1 0

0 1 2 3 4 5 6 7
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
3 4 5 6 7 8 9 10
4 5 6 7 8 9 10 11
5 6 7 8 9 10 11 12
6 7 8 9 10 11 12 13
7 8 9 10 11 12 13 14

As long as you judge I-t==a[j]-j and T+i==a[j]+j, then proceed to the next column.

#include <stdio.h>inta[8];intCnt=0;voidEtinta[8],intt) {    inti,j; if(t==8) cnt++; Else    {         for(i=0; i<8; i++)        {            intok=1; A[t]=i;  for(j=0; j<t; J + +)            {                if(a[j]==i| | i-t==a[j]-j| | t+i==a[j]+j) {OK=0;  Break; }            }            if(OK) et (a,t+1); }    }}intMain () {ET (A,0); printf ("%d", CNT);}

4. Two points Search

Press semi-search to

#include <stdio.h>a[Ten]={6,8, -, -, $, $, the, -, About,102};intMain () {intErintTintUpintDown ); intt,ant; scanf ("%d",&t); Ant=er (T,9,0); printf ("%d", ant);}intErintTintUpintDown ) {    intm;  while(up!=Down ) {m=down+ (Up-down)/2; if(a[m]==t)returnm+1; if(a[m]<t) down=m+1; Else up=m; }    return-1;}

5. Enumeration arrangement

According to the eight Queen's method, the following

A[1000] is used to save the result, the IF statement within the PL function determines whether it is marginal, else is used to loop 1 to N and backtrack.

#include <stdio.h>inta[ +];intN;intMain () {voidPlintTinta[ +]); scanf ("%d",&N); PL (0, a);}voidPlintTinta[ +]){    inti,j; if(t==N) { for(i=0; i<n;i++) printf ("%d", A[i]); printf ("\ n"); }    Else{         for(j=1; j<=n;j++)        {            intK; intok=1;  for(k=0; k<t;k++)            {                if(A[K]==J) {ok=0; Break;} }            if(OK) {A[T]=J;PL (t+1, a);} }    }}

6, Match stick problem

Direct enumeration, first enumerate the first two, calculate the third, judge the three required match stick number of sum plus 4 is equal to N, can.

#include <stdio.h>intnum[]={6,2,5,5,4,5,6,3,7,6};intCnt=0;intDeintd) {    intsum=0; if(d==0)return 6;  while(d) {sum+=num[d%Ten]; D=d/Ten; }    returnsum;}intMain () {intN; inti,j; scanf ("%d",&N);  for(i=0;i< +; i++)    {         for(j=0;j< +; j + +)        {            if(De (i) +de (j) +de (I+J) +4==N) CNT++; }} printf ("%d", CNT);}

7

Recursion, enumeration