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HDU1250 Hat's Fibonacci 【亿进制】
Hat‘s Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7854 Accepted Submission(s): 2551
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.
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
4203968145672990846840663646 Note: No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.
#include <stdio.h> #include <string.h> #define mod 1000000000 int arr[10001][230]; void Add(int i, int j) { for(int k = 0; k < 230; ++k) { arr[i][k] += arr[j][k]; if(arr[i][k] >= mod) { arr[i][k] -= mod; ++arr[i][k+1]; } } } int main() { int i, j, n; for(i = 1; i < 5; ++i) arr[i][0] = 1; for(i = 5; i < 10001; ++i) { for(j = i - 4; j < i; ++j) Add(i, j); } while(scanf("%d", &n) == 1) { for(j = 229; ; --j) if(arr[n][j]) { printf("%d", arr[n][j]); break; } for(--j; j >= 0; --j) printf("%09d", arr[n][j]); printf("\n"); } return 0; }
HDU1250 Hat's Fibonacci 【亿进制】
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