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Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12650????Accepted Submission(s): 4253


Problem Description A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.

Input Each line will contain an integers. Process to end of file.

Output For each case, output the result in a line.
Sample Input 100 Sample Output 4203968145672990846840663646Note:No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.
Author 戴帽子的
Recommend Ignatius.L???|???We have carefully selected several similar problems for you:??1753?1865?1715?1002?2100

解法：

題中要求的數(shù)最大可能到兩千多位，直接開二維數(shù)組會空間超限，
所以可以用a[ ][ ]存取一個四位，五位…或八位數(shù)等，減小數(shù)組的大小，
下面以五位數(shù)為例，每次算的結(jié)果由以前的對10取余改變成對10萬取余，
輸出的時候找到第一位不是0的輸出，以后的都按照%05d輸出

#include<stdio.h> #include<string.h> int a[10001][520]; void fun() {a[1][0]=1;a[2][0]=1;a[3][0]=1;a[4][0]=1;for(int i=5; i<10000; i++){int t=0;for(int j=0; j<=510; j++){a[i][j]=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j]+t;t=a[i][j]/100000;a[i][j]=a[i][j]%100000;}}} int main() {fun();int n;while(~scanf("%d",&n)){int i;for( i=510; a[n][i]==0; i--);printf("%d",a[n][i]);for(int j=i-1; j>=0; j--)printf("%05d",a[n][j]);printf("\n");}return 0; }

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