大斐波数
大斐波数
题目: Fibonacci数列,定义如下: f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3。 计算第n项Fibonacci数值。 Input 输入第一行为一个整数N,接下来N行为整数Pi(1<=Pi<=1000)。 ......
http://acm.hdu.edu.cn/ 杭电 1465 1466 1480 1555 1597 1702 1715 1720 1725 1727 1732 1735 1797 1803 1804 1859 1860 acm的AC解题报告
大菲波数 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 21198 Accepted Submission(s): 7163 Problem Description ...Fibonacci数列,定义如下:...
/**/ #include &lt;cstdio&gt; #include &lt;cstring&gt; #include &lt;cmath&gt; #include &lt;cctype&gt; #include &lt;iostream&gt; #include &...set&
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http://acm.hdu.edu.cn/showproblem.php?pid=1715 大菲波数 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22523 Accepted Submission(s): ...
Problem Description Fibonacci数列,定义如下: f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3。 计算第n项Fibonacci数值。 Input 输入第一行为一个整数N,接下来N行为整数Pi(1 ...Sample
大菲波数Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22722 Accepted Submission(s): 8163Problem DescriptionFibonacci数列,定义如下:f(1)=f(2)=1...
大菲波数Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22722 Accepted Submission(s): 8163Problem DescriptionFibonacci数列,定义如下:f(1)=f(2)=1...
标签: hdu
题目链接 大菲波数 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 23754 Accepted Submission(s): 8517 Problem Description ...
#include #include using namespace std; string add(string s1,string s2) { if(s1==""&&s2=="") return "0"; if(s1=="") return s2; if(s2=="") return s1; int j,l,la,lb;... stri
import java.math.BigDecimal; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int T = sc.nextInt();...
/* 中文题意: 中文翻译: 题目大意: 解题思路: 难点详解: 关键点: 解题人:lingnichong 解题时间:2014/7/31 20:42 解题感受: ...Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/...
Problem Description Fibonacci数列,定义如下: f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3。 计算第n项Fibonacci数值。 Input 输入第一行为一个整数N,接下来N行为整数Pi(1<=Pi<=1000)。...
Fibonacci数列,定义如下: f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3。 计算第n项Fibonacci数值。 Input 输入第一行为一个整数N,接下来N行为整数Pi(1 Output 输出为N行,每行为对应的f...Sampl
题目大意:输入一个整数t表示测试用例数。接着输入n,输出对应的斐波那契数fib[n]。(一开始,我看成了输入n,然后输出前n个斐波那契数。。。蛋碎了一地) ...解题思路:对于用大数来解决的斐波那契数列的相关题目。...
原题链接 import java.util.Scanner; import java.math.BigInteger; public class Main{ public static void main(String[] args){ final int max = 1001;... Scanner cin = new Scanner(System.in);...
大菲波数 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9224 Accepted Submission(s): 3133 Problem Description Fibonacci数列,定义如下: ...
大菲波数 Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7677Accepted Submission(s): 2557 Problem Description Fibonacci数列,定义如下: f(1)=f....
import java.io.*; import java.math.*; import java.util.*; public class Main { /** * @param args */ public static void main(String[] args) { // TODO 自动生成的方法... Scanner cin = ...
import java.math.*; import java.util.*; import java.io.*; public class Main { public static void main(String[] args) { int n,p,i; BigInteger sum[]=new BigInteger[1005];... sum[1]=BigI...
大菲波数 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 26717 Accepted Submission(s): 9592 Problem Description Fibonacci数列,定义如下: ...
ac代码: #include<stdio.h> int a[1006][504] ={0}; int main() { int n, m, i, j, k; a[1][0] = 1; a[2][0] = 1; for(i = 3; i <= 1000; i++) { k = 0; ... k = a[i]