How Many Fibs?
2015-08-05 13:37
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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:
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.csdn.net/blog/u013451221/29792437
参考链接:http://m.blog.csdn.net/blog/u013451221/29792437
参考链接:http://m.blog.csdn.net/blog/u013451221/29792437
谢谢
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.csdn.net/blog/u013451221/29792437
参考链接:http://m.blog.csdn.net/blog/u013451221/29792437
参考链接:http://m.blog.csdn.net/blog/u013451221/29792437
谢谢
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