HDU 1316 How Many Fibs? 大数

HDU 1316 题目链接

How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3281 Accepted Submission(s): 1312

[align=left]Problem Description[/align]
Recall the definition of the Fibonacci numbers:

f1 := 1

f2 := 2

fn := fn-1 + fn-2 (n >= 3)

Given two numbers a and b, calculate how many Fibonacci numbers are in the range [a, b].

[align=left]Input[/align]
The input contains several test cases. Each test case consists of two non-negative integer numbers a and b. Input is terminated by a = b = 0. Otherwise, a <= b <= 10^100. The numbers a and b are given with no superfluous leading zeros.

[align=left]Output[/align]
For each test case output on a single line the number of Fibonacci numbers fi with a <= fi <= b.

[align=left]Sample Input[/align]

10 100
1234567890 9876543210
0 0

[align=left]Sample Output[/align]

5
4

[align=left]Source[/align]
University of Ulm Local Contest 2000

[align=left]Recommend[/align]
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好吧，这道题目，让我WA了3次，实在是想哭，最后又重看了一遍题目，发现 f[1] = 1 f[2] = 2 ， 惯性思维害死人

思路也没什么特别的 大数相加，然后大数的比较

值得一提的还是审题啊

#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;

struct ds{
int num[105] ;
int len ;
void init( char s[105] = "0" ){
memset( num , 0 , sizeof(num) );
len = 0 ;
int l = strlen(s) ;
for( ; len < l ; ++len ){
num[len] = s[l-1-len] - '0' ;
}
return ;
}
void plus( const ds& a , const ds& b ){
this->init();
int maxlen = a.len>b.len?a.len:b.len;
for( len = 0 ; len < maxlen ; ++len )
num[len] = a.num[len] + b.num[len] ;
for( len = 0 ; len < maxlen ; ++len ){
num[len+1] += num[len]/10 ;
num[len] %= 10 ;
}
if( num[len] )
++len ;
return ;
}
bool is_less_than_or_equal_to( const ds& a ){
if( len != a.len )
return len<a.len ;
for( int i = len - 1 ; i >= 0 ; --i ){
if( num[i] != a.num[i] )
return num[i]<a.num[i] ;
}
return true;
}
}fb[500];

int main(){
fb[1].init("1");
fb[2].init("2");
int i , ans ;
for( i = 3 ; i < 500 ; ++i )
fb[i].plus(fb[i-1],fb[i-2]);
char a[105] , b[105] ;
ds aa , bb ;
while( scanf( "%s%s" , a , b ) ){
if( strcmp(a,"0") == 0 && strcmp(b,"0") == 0 )
break;
aa.init(a);
bb.init(b);
ans = 0;
for( i = 0 ; i < 500 ; ++i )
if( aa.is_less_than_or_equal_to(fb[i]) && fb[i].is_less_than_or_equal_to(bb) )
++ans;
printf( "%d\n" , ans );
}
return 0;
}