Java
import java.io.*;import java.util.*;import java.math.*;public class aa{ public static void main( String[] args ) { BigInteger f[] = new BigInteger[505],a,b; f[1] = BigInteger.valueOf(1); f[2] = BigInteger.valueOf(2); for( int i = 3; i < 505; ++i ) f[i] = f[i-1].add( f[i-2] ); Scanner cin = new Scanner(System.in); while( cin.hasNextBigInteger() ) { a = cin.nextBigInteger(); b = cin.nextBigInteger(); if( a.compareTo(BigInteger.ZERO) == 0 && b.compareTo(BigInteger.ZERO) == 0) break; int c = 0; for( int i = 1; i < 505; ++i ) { if( f[i].compareTo(b) > 0 ) break; if( f[i].compareTo(a) >= 0 && f[i].compareTo(b) <= 0 ) ++c; } System.out.println( c ); } }} 这题很明显只能用大数,先把fibs求出来,然后再去找区间
#include#include #include #define max 124char str1[max],str2[max];char ch[10000][max] ={0};void chart( ){ int i = 3,len = 0; ch[1][0] = '1'; ch[2][0] = '2'; for( ; len < max; ++i ) { int j = 0, t = 0; while( ch[i - 2][j] ) { t += ch[i-2][j] - '0' + ch[i-1][j] -'0'; ch[i][j] = t % 10 + '0'; t /= 10; ++j; } while( ch[i-1][j] ) { t += ch[i-1][j] - '0'; ch[i][j] = t % 10 + '0'; t /= 10; ++j; } while( t ) { ch[i][j] = t % 10 + '0'; t /= 10; ++j; } len = strlen( ch[i] ); } }int cmp( char str[],char str2[] ){ char str1[max] = {0};//别丢了这一步,否则WA strcpy( str1,str ); int len1 = strlen( str1 ),len2 = strlen( str2 ); if( len1 > len2 ) return 1; else if( len1 < len2 ) return -1; else { for( int p = len1 - 1,q = 0; p > q;--p,++q ) { char c = str1[p]; str1[p] = str1[q]; str1[q] = c; } return strcmp( str1 ,str2 ); }}int cal( ){ int sum = 0,i = -1; while( cmp(ch[++i],str1) < 0 ); while( cmp( ch[i],str2 ) <= 0 ) { ++i; ++sum; } return sum;}int main( ){ chart( ); while( scanf( "%s%s",str1,str2 ),str1[0] + str2[0] != '0' + '0' ) printf( "%d\n",cal( ) ); return 0;}