杭电1250Hat's Fibonacci

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9134    Accepted Submission(s): 2970

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

妥妥的水题，但是要是存二维数组，第二维只存一位数的话会超内存的，超了3次才想到存4位数；所以第二维只需要开到2005/4即可；一维开到8000即可.

#include<stdio.h>#include<string.h>#include<algorithm>#define N 2010int a[7061][555];int i,n,m,j,k,l;int main(){memset(a,0,sizeof(a)); a[1][0]=a[2][0]=a[3][0]=a[0][0]=1; for(i=4;i<7060;i++) { k=0; for(j=0;j<555;j++) { a[i][j]=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j]+k; k=a[i][j]/10000; a[i][j]=a[i][j]%10000; } } while(scanf("%d",&n)!=EOF) { j=550; while(a[n-1][j]==0)j--;printf("%d", a[n-1][j--]);//这一位不一定是四位数，所以输出%d for(; j>=0; j--)printf("%04d", a[n-1][j]);//剩下的一定是4位数的，所以输出%04d，否则就会错.！！！！！！ putchar('\n');//printf("%d",a[n-1][j]);//这的输出也是ac代码； //i=j-1;//个人比较喜欢下边的输出 //for(i=j-1;i>=0;i--)//printf("%04d",a[n-1][i]);//printf("\n"); }return 0;}