[求助]经典兔子问题

古典问题：有一对兔子，从出生后第3个月起每个月都生一对兔子，小兔子长到第三个月
　　　后每个月又生一对兔子，假如兔子都不死，问每个月的兔子总数为多少？
那位好心人帮俺写上，俺想了好几天没想出来
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去网上搜"经典程序百例"
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#include "stdio.h"
#include "conio.h"
main()
{
  long f1,f2;
  int i;
  f1=f2=1;
  for(i=1;i<=20;i++)
  {
    printf("%12ld %12ld",f1,f2);
    if(i%2==0) printf("\n"); /*控制输出，每行四个*/
    f1=f1+f2; /*前两个月加起来赋值给第三个月*/
    f2=f1+f2; /*前两个月加起来赋值给第三个月*/
  }
  getch();
}
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上面好像没有体现"出生后第三个月起"的意思啊???
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#include<iostream.h> void main(){ int months,i; long int firstmonth=1,secondmonth=0,adult=0; cin>>months; for(i=1;i<=months;i++){ cout<<"the "<<i<<"month ,the number of rubbit is"<<adult<<endl;

adult=adult+secondmonth; secondmonth=firstmonth; firstmonth=adult; cout<<"the total is "<<adult<<endl; }; }

不知行不？

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呵呵我做出来了,还做了张图
#include &lt;stdio.h&gt;
#include &lt;stdlib.h&gt;
#include &lt;graphics.h&gt;

void main(void)
{
    int sum,r1=1,r2=0,r3=0;
    int i,j,k;
    int n;
    int g=VGA,mode=VGAHI;
int x,y;
    int x0,y0;
    int *p;
    clrscr();
    printf("input the n:");
    scanf("%d",&amp;n);
    p=(int *)malloc(n*sizeof(int));
    if (p==NULL)
    {
        printf("\nwrong!\n");
        exit(0);
    }
    printf("input the 3p:");
    scanf("%f%f%f",&amp;p1,&amp;p2,&amp;p3);
    initgraph(&amp;g,&amp;mode,"");
    setbkcolor(WHITE);
    setcolor(BLUE);
    line(150,380+5,150,50);
    line(150,380,500,380);
    moveto(150,50);
    linerel(-3,6);
    moveto(150,50);
    linerel(3,6);
    moveto(500,380);
    linerel(-6,-3);
    moveto(500,380);
    linerel(-6,3);
    for (j=0;j&lt;=300;j+=25)
    {
        line(145,380-j,150,380-j);
    }
    for (k=0;k&lt;=300;k+=25)
    {
        line(k+150,380,k+150,377);
    }
    for (i=2;i&lt;=n;i++)
    {
        r3=r3+;
        r2=r1;
        r1=r3;
        sum=r1+r2+r3;
        x0=i*10;
        y0=sum/2;
        x=x0+150;
        y=480-1*y0-100;
        *(p+(i-2)*2)=x;
        *(p+(i-2)*2+1)=y;
    }
    drawpoly(n,p);
    getchar();
    getchar();
    getchar();
    closegraph();
free(p);
}
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main()
{
int i, ny,sum,n=3;
printf("please input the number of \'ny\':");
scanf("%d",&amp;ny);
if(ny&gt;=3)
{
  sum=ny-3+2;
  for(;n&lt;ny;n++)
  {
   i=1;
   while((ny-n)&gt;=3*i)
   {
    sum+=ny-n-3*i+1;
    i++;
   }
  }
}
else sum=1;
printf("%d\n",sum);
}
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老谭那书里有这题啊

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