杭电1316

来源：互联网发布：javascript排序编辑：程序博客网时间：2024/06/18 13:55

题目描述：

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].

求斐波那契数列在[a,b]区间的个数，这题弄得累死了，直接上AC代码：

#include<stdio.h>#include<string.h>int a[10000][300]={0};int b[10000];char ch[102];int compare(int a[],char b[],int n){for(int i=0;i<n;i++)ch[i]=a[n-1-i]+'0';ch[n]='\0';return strcmp(ch,b);}int main(){a[0][0]=1;a[1][0]=1;b[0]=1;b[1]=1;int i;for(i=2;i<1000;i++){int n=b[i-2],m=b[i-1],flag=0,j;for(j=0;j<n;j++){int t=a[i-2][j]+a[i-1][j]+flag;if(t>9){flag=1;a[i][j]=t-10;}else {a[i][j]=t;flag=0;}}for(;j<m;j++){int t=a[i-1][j]+flag;if(t>9){flag=1;a[i][j]=t-10;}else {a[i][j]=t;flag=0;}}if(flag==1) a[i][m++]=1;b[i]=m;a[0][0]=-1;}char ch1[102],ch2[102];while(scanf("%s %s",&ch1,&ch2)!=EOF){int l1=strlen(ch1);int l2=strlen(ch2);if(l1==1 && l2==1 && ch1[0]=='0' && ch2[0]=='0') break;int q,k,r;for(q=0;q<999;q++)if(b[q]>l1-1) break;for(i=0;i<999;i++)if(b[i]>l1) break;for(k=q;k<i;k++){if(compare(a[k],ch1,l1)>0) break;}        if(compare(a[k],ch1,l1)==0) k++;for(q=0;q<999;q++)if(b[q]>l2-1) break;for(i=0;i<999;i++)if(b[i]>l2) break;for(r=q;r<i;r++){if(compare(a[r],ch2,l2)>0) break;}if(compare(a[r],ch2,l2)==0) r++;if(r>k){if(compare(a[k-1],ch1,l1)==0 && compare(a[r-1],ch2,l2)<=0) printf("%d\n",r-k+1);else printf("%d\n",r-k);}else{if(compare(a[r-1],ch1,l1)==0) printf("1\n");else printf("0\n");}}return 0;}

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