How Many Fibs? (大数)
发布时间:2020-12-14 03:38:21 所属栏目:大数据 来源:网络整理
导读:How Many Fibs? Time Limit : 2000/1000ms (Java/Other)???Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 15???Accepted Submission(s) : 6 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Recall the
How Many Fibs?Time Limit : 2000/1000ms (Java/Other)???Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 15???Accepted Submission(s) : 6Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description
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]. Input
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.
Output
For each test case output on a single line the number of Fibonacci numbers fi with a <= fi <= b.
Sample Input10 100 1234567890 9876543210 0 0 Sample Output5 4 Source
University of Ulm Local Contest 2000
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题意:就是给出a,b;(a<b)求出a,b之间的斐波那契数的个数;
思路:先求出1-500的斐波那契数(因为第500个斐波那契数就已经超过100位了)
代码:
#include<iostream> #include<string.h> #include<cstdio> using namespace std; char f[501][1000]; void add(char a[],char b[],char s[])//大数加法函数(斐波那契数用大数加法来做的) { int i,j,k,up,x,y,z,l; char *c; if (strlen(a)>strlen(b)) l=strlen(a)+2; else l=strlen(b)+2; c=new char[l]; i=strlen(a)-1; j=strlen(b)-1; k=0; up=0; while(i>=0||j>=0) { if(i<0) x='0'; else x=a[i]; if(j<0) y='0'; else y=b[j]; z=x-'0'+y-'0'; if(up) z+=1; if(z>9) { up=1; z%=10; } else up=0; c[k++]=z+'0'; i--; j--; } if(up) c[k++]='1'; i=0; c[k]=' '; for(k-=1; k>=0; k--) s[i++]=c[k]; s[i]=' '; } int cmp(char a[],char b[])//比较函数就是先看长度,再看字符(和strcmp函数差不多)相等时返回0,大于时,返回正数,小于时,返回负数。 { int i,len1,len2; len1=strlen(a); len2=strlen(b); if(len1>len2) return 1; else if(len1<len2) return -1; else if(len1==len2) { for(i=0;i<len1;i++) { if(a[i]!=b[i]) { break; } } return a[i]-b[i]; } } int main() { int i,len2,m,t,s; char a[102],b[102]; f[0][0]='1'; f[1][0]='1'; for(i=2;i<501;i++) { add(f[i-1],f[i-2],f[i]); } while((scanf("%s%s",a,b)!=EOF)&&!(a[0]=='0'&&b[0]=='0')) { m=0; len1=strlen(a); len2=strlen(b); for(i=1;i<501;i++) { t=cmp(a,f[i]); s=cmp(b,f[i]); if((t<0||(t==0))&&(s>0||s==0))//判断斐波那契数是否在a,b之间,再,就计数; m++; } cout<<m<<endl; } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |