How Many Fibs?

最新推荐文章于 2021-04-18 09:15:45 发布

OVS98

最新推荐文章于 2021-04-18 09:15:45 发布

阅读量623

点赞数

分类专栏： ACM 文章标签：大数

ACM 专栏收录该内容

37 篇文章

订阅专栏

题目来自杭电：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): 5214 Accepted Submission(s): 2032

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
10 100
1234567890 9876543210
0 0

Sample Output
5
4

Source
University of Ulm Local Contest 2000
代码：

#include <iostream>
#include <cstring>
using namespace std;
const int MAX = 1000;
const int LEN = 100 + 10;
int aFib[MAX][LEN];
//相加
int add(int *x,int *y,int *z){
    int i;
    int temp;
    int carry = 0;

    for(i = 0;i < LEN;i++){
        temp = x[i] + y[i] + carry;
        z[i] = temp % 10;
        carry = temp /10;
    }
    return 0;
}

//获得表
int Fib(){
    int i;
    memset(aFib,0,sizeof(aFib));

    aFib[0][0] = 1;
    aFib[1][0] = 2;
    for(i = 2;i < MAX;i++)
        add(aFib[i-1],aFib[i-2],aFib[i]);
    return 0;
}
//比较
int compare(char *x,int *y){
    int i,j;
    int temp[LEN] = {0};
    for(i = 0,j = strlen(x) - 1;j >= 0;i++,j--)
        temp[i] = x[j] - '0';

     for(i = LEN - 1;i >= 0;i--){
        if(temp[i] > y[i])
            return 0;
        else if(temp[i] < y[i])
            return 1;
     }
     return 2;
}
int main()
{
    char a[LEN],b[LEN];
    int head,tail;
    int i,j;
    int temp;
    Fib();

    while(cin >> a >> b){
        if(strcmp(a,"0") == 0 && strcmp(b,"0") == 0)
            break;
        for(i = 0;i < MAX;i++){
            temp = compare(a,aFib[i]);
            if(temp){
                head = i;
                break;
            }
        }
        for(;i < MAX;i++){
            temp = compare(b,aFib[i]);
            if(temp == 1){
                tail = i;
                break;
            }
            else if(temp == 2){
                tail = i + 1;
                break;
            }
        }
        cout << tail - head << endl;
    }
    return 0;
}

小结：大数问题，思想是先建立斐波那契表，然后比对，通过数组下标计算个数，建立斐波那契数表的过程也就是大数相加的过程，这个不难，巧妙地是统计个数，原来我也用到了vector,一个一个比较，虽然AC了，但是效率很低，然后看了别人的代码，灵光一闪，只要比较首尾就可以了，然后通过下标技术就可以了，于是又重写了别人的代码，基本上没变；这种方法比我之前的效率高很多
参考链接：http://m.blog.youkuaiyun.com/blog/u013451221/29792437
参考链接：http://m.blog.youkuaiyun.com/blog/u013451221/29792437
参考链接：http://m.blog.youkuaiyun.com/blog/u013451221/29792437
谢谢