一对兔子 长到3个月成熟，3个月后每个月生一对兔子。生下的小兔子也符合这个条件。问：输入月份求出该月份的兔子总数。

#include<stdio.h>

void getmoon(int n)/*n为共需输出几个月的数*/
{
    int i=0,s,sum,x=1,y1=0,y2=0,y3=0;/*x用来存成年的，y用来存小的,i用来存月数*/
    for(i=1;i<=n;i++)
    {
        s=y3; /*3月龄的现再也应成年了，下面行同理*/
        y3=y2;
        y2=y1;
        y1=x*2; /*这个月新生小免数*/
        x+=s;  /*这月底的成年数，下月要生小免的*/
      
        sum=y3+y2+y1+x;
      
        printf("i moonth: %d\n",sum);
      
     
    }

 

 
}

main()
{
    int sum;
    printf("Please enter a number:");
    scanf("%d",&sum);
    getmoon(sum);

 
}

先吃饭 回来再研究 谢谢了

你得出的怎么都是奇数? 一对兔子生1对小兔子 怎么算也是双数才对不是吗

#include"stdio.h"
#include"conio.h"
int getmoon(int n)/*n为共需输入几个月的数*/
{
    int y3=1,y2=0,y1=0,x=0,i,sum,p3,p2,p1;/*y3表示差三月成熟的，y2表示差两月成熟的，y1表示差一月成熟的，x表示已经成熟的。  */
        for(i=1;i<=n;n++)
        {
            if(i==1)
            {
                y3--;y2++;
            }
            else if(i==2)
                    {
                        y2--;
                        y1++;
                    }
                else if(i==3)
                        {
                            y1--;
                            x++;
                        }
                    else if(i==4)
                            {
                                y3++;
                            }
                        else if(i==5)
                                {
                                    y2++;
                                }
                            else if(i==6)
                                    {
                                        y1++;
                                    }
                                else if(i>=7)
                                        {
                                            p3=y3;
                                            y3=x;
                                            p2=y2;
                                            y2=p3;
                                            p1=y1;
                                            y1=p2;
                                            x=x+p1;
                                        }
     }
    sum=(y1+y2+y3+x);
    return sum;
}
 
void main()
 {
     int n,sum;
     printf("Please enter a number:\n");
     scanf("%d",&n);
     sum=getmoon(n);
     printf("%d",sum);
     printf("\n");
     getch();
 }
 

路过学习

这个问题注意，从第三个月起  就满足A3=A2+A1;就是后一个数等于前两个数之和，如果你注意到这点了，这个问题就好说了，我也是上几周把这个程序编写出来的。。。。

#include <stdio.h>
#include <stdlib.h>
struct Month_rabbit
{
    long int
        old_rabbit,   //成熟的兔子
        new_rabbit,   //新生的兔子
        unew_rabbit,  //非新生未成熟的兔子
        sum_rabbit;   //该月为止兔子总数
} ;

void rabbit_sum(int month, struct Month_rabbit *prabbit)
{
    if(month<1)
    {
        printf("输入有误！"); exit(1);
    }

    for(int i=1;i <= month; i++ )
    {
        switch(i)
        {
            case 1 :
                prabbit->new_rabbit= 2; prabbit->old_rabbit = 0; prabbit->unew_rabbit = 0; prabbit->sum_rabbit = 2;
            break;
            case 2 :
                prabbit->new_rabbit= 0; prabbit->old_rabbit = 0; prabbit->unew_rabbit = 2; prabbit->sum_rabbit = 2;
            break;
            case 3 :
                prabbit->new_rabbit= 0; prabbit->old_rabbit = 0; prabbit->unew_rabbit = 2; prabbit->sum_rabbit = 2;
            break;
            default :
                prabbit->old_rabbit = (prabbit-3)->new_rabbit + (prabbit-1)->old_rabbit ;//本月成熟兔子等于往前数三个月月份新生兔子与本月前一个月月份成熟兔子之和
                prabbit->unew_rabbit = (prabbit-1)->unew_rabbit + (prabbit-1)->new_rabbit - (prabbit-3)->new_rabbit ;//本月非新生未成熟兔子等于前一个月份非新生未成熟兔子加前一个月份新生兔子的和减去本月往前数三个月月份的新生兔子数
                prabbit->new_rabbit = prabbit->old_rabbit ;//本月新生兔子数等于本月成熟的兔子数
                prabbit->sum_rabbit = prabbit->old_rabbit + prabbit->new_rabbit + prabbit->unew_rabbit ;//本月为止兔子总数为本月成熟的兔子数和新生的兔子数以及非新生未成熟兔子数之和
      
        }
        prabbit++;
    }
}

int main(void)
{
    int month;
    long int total = 0;
    struct Month_rabbit rabbit[20];
    printf("请输入月份：");
    scanf("%d", &month);
    rabbit_sum(month, rabbit);
    printf("到第%d月为止共有兔子：%ld\n", month,rabbit[month-1].sum_rabbit);
    return 0;
}

[ 本帖最后由 liao06550107 于 2011-11-1 18:42 编辑 ]

都谢谢了 恩

/* Note:Your choice is C IDE */
#include "stdio.h"
void main()
{
    int yue,i,c;
    int m=1,n=1,sum;
    printf("请输入过了几个月：");
    scanf("%d",&yue);
    if(yue==1||yue==2)
    {
        printf("过了%d兔子的数量变为1\n",yue);
    }
    else
    {
        
        for(i=3;i<yue+1;i++)
        {
             sum=m+n;
              n=m;
             m=sum;
        }
   
        printf("过了%d月兔子的数量为%d",yue,sum);   
    }   
}

试试这个，应该就是这个了。我只是觉得上面的太复杂了，所以就发这个了。
你试试