HDU 1316 (斐波那契数列,大数相加,大数比较大小)
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题目链接:
http://acm.hdu.edu.cn/showproblem.php?pid=1316
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 Input
10 100
1234567890 9876543210
0 0
Sample Output
5
4
代码如下:
#include<bits/stdc++.h> using namespace std; string str[500]; string add(string str1,string str2)//大数相加 { int l1=str1.length(); int l2=str2.length(); if(l1>l2) { for(int i=0;i<l1-l2;i++) { str2="0"+str2; } }else if(l1<l2) { for(int i=0;i<l2-l1;i++) { str1="0"+str1; } } l1=str1.length(); string str3; int c=0; for(int i=l1-1;i>=0;i--) { int temp=str1[i]-‘0‘+str2[i]-‘0‘+c; c=temp/10; temp=temp%10; str3=char(temp+‘0‘)+str3; } if(c!=0) { str3=char(c+‘0‘)+str3; } return str3; } int compare(string str1,string str2)//str1大于等于str2,返回1 { int l1=str1.length(); int l2=str2.length(); if(l1>l2) { return 1; }else if(l1<l2) { return 0; }else { for(int i=0;i<l1;i++) { if(str1[i]>str2[i]) { return 1; }else if(str1[i]==str2[i]) { continue; }else { return 0; } } } return 1; } int f(string str1,string str2)//找出两个大数中间的斐波那契数的个数,包括边界 { int l1=str1.length(),l2=str2.length(),index1,index2; for(int i=0;i<=500;i++) { int k=str[i].length(); if(k<l1) continue; else { index1=i; break; } } for(int i=499;i>=0;i--) { int k=str[i].length(); if(k>l2) continue; else { index2=i; break; } } int r=0; for(int i=index1;i<=index2;i++) { if(compare(str[i],str1)==1&&compare(str2,str[i])==1) { r++; } } return r; } int main() { string str1,str2; str[0]="1"; str[1]="2"; for(int i=2;i<500;i++)//先求出需要的斐波那契数列,第500个斐波那契数大于10的1000次方 { str[i]=add(str[i-1],str[i-2]); } while(cin>>str1>>str2) { if(str1=="0"&&str2=="0") break; int r=f(str1,str2); printf("%d\n",r); } return 0; }
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