hdu1316 大数

题意:

     给你一个区间,问这个区间有多少个斐波那契数.

思路:

     水的大数,可以直接模拟,要是懒可以用JAVA,我模拟的,打表打到1000个就足够用了...

#include<stdio.h>
#include<string.h>

int a[110] ,b[110];
int num[1200][110];

void csh_num()
{
   memset(num ,0 ,sizeof(num));
   num[1][1] = 1 ,num[1][0] = 1;
   num[2][1] = 2 ,num[2][0] = 1;
   for(int i = 3 ;i <= 1000 ;i ++)
   {
      for(int j = 1 ;j <= 105 ;j ++)
      num[i][j] = num[i-1][j] + num[i-2][j];
      for(int j = 1 ;j <= 105 ;j ++)
      {
         num[i][j+1] += num[i][j] / 10;
         num[i][j] %= 10;
      }
   }

   for(int i = 3 ;i <= 1000 ;i ++)
   {
      int kg = 0 ,sum = 0;
      for(int j = 105 ;j >= 1 ;j --)
      {
         if(num[i][j]) kg = 1;
         if(kg) num[i][0]++;
      }
   }
   return ;
}

bool campa(int i ,int la)
{
   if(num[i][0] > la) return 1;
   if(num[i][0] < la) return 0;
   for(int ii = la ;ii >= 1 ;ii --)
   {
      if(num[i][ii] == a[ii]) continue;
      if(num[i][ii] > a[ii]) return 1;
      else return 0;
   }
   return 1;
}

bool campb(int i ,int lb)
{
   if(num[i][0] > lb) return 0;
   if(num[i][0] < lb) return 1;
   for(int ii = lb ;ii >= 1 ;ii --)
   {
      if(num[i][ii] == b[ii]) continue;
      if(num[i][ii] > b[ii]) return 0;
      else return 1;
   }
   return 1;
}

int main ()
{
   csh_num();
   int i ,j ,sum;
   char stra[105] ,strb[105];
   while(~scanf("%s %s" ,stra ,strb) && strcmp(stra ,"0") + strcmp(strb ,"0"))
   {
      memset(a ,0 ,sizeof(a));
      int la = strlen(stra) - 1;
      int tt = 0;
      for(i = la ;i >= 0 ;i --)
      a[++tt] = stra[i] - 48;

      memset(b ,0 ,sizeof(b));
      int lb = strlen(strb) - 1;
      tt = 0;
      for(i = lb ;i >= 0 ;i --)
      b[++tt] = strb[i] - 48;
      la++ ,lb++;

      sum = 0;
      for(i = 1 ;i <= 1000 ;i ++)
      {
         if(!campb(i ,lb)) break;
         if(campa(i ,la)) sum ++;
      }

      printf("%d\n" ,sum);
   }
   return 0;
}