hdu1316 How Many Fibs?（大数）

http://acm.hdu.edu.cn/showproblem.php?pid=1316

How Many Fibs?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1451    Accepted Submission(s): 604

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 Input

10 1001234567890 98765432100 0

 

Sample Output

54

 

 

code：

#include<iostream>#include<string>#include<cmath>using namespace std;#define MaxSize 110#define N 490struct BigN{int a[MaxSize];int len; //保存大数的位数}f[N];//打印数列void start(){int i, j, ch;f[1].a[1] = 1;f[2].a[1] = 2;f[2].len = f[1].len = 1;for(i = 3; i < N; i++){ch = 0;for(j = 1; j <= f[i-2].len; j++){f[i].a[j] = f[i-1].a[j] + f[i-2].a[j] + ch;ch = f[i].a[j] / 10;if(f[i].a[j] > 9)f[i].a[j] %= 10;}//较大的数的第一位单独考虑if(f[i-1].len > f[i-2].len){f[i].a[j] = f[i-1].a[j] + ch;ch = f[i].a[j] / 10;if(ch > 0)f[i].a[j] %= 10;j++;}//仍有进位时，保存进位if(ch > 0)f[i].a[j++] = ch;f[i].len = j - 1;}}//字符串转整型void CharToInt(string s, int *a){int i;for(i = 1; i < s.length() + 1; i++){a[i] = s[s.length() - i] - '0';}}//比较两大数的大小int com(int *a, int n){int i,t;for(i = f[n].len; i>=1; i--){t = f[n].a[i] - a[i];if(t > 0) return 1;if(t < 0)return -1;}return 0;}int main(){int i;string s1, s2;start();while(cin>>s1>>s2){if(s1 == "0" && s2 == "0") break;int len1 = s1.length();int len2 = s2.length();int *a = new int[len1+2];int *b = new int[len2+2];CharToInt(s1,a);CharToInt(s2,b);int t1,t2,t3,t4,t,t0;t1 = t2 = t3 = t4 = 0;//保存与两目标大数长度相同的数的位置for(i = 1; i <= N ; i++){t = f[i].len - len1;t0 = f[i].len - len2;if(!t && !t1) t1 = i;if(t == 1 && !t2) t2 = i;if(!t0 && !t3) t3 = i;if(t0 == 1 && !t4) {t4 = i;break;}}int n = t4 - t1;//除去位数相同大小符合条件的数for(i = t1; i < t2; i++){t = com(a,i);if(t == -1)n--;}for(i = t3; i < t4; i++){t = com(b,i);if(t == 1) n--;}cout<<n<<endl;delete[] a;delete[] b;}return 0;}