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HDU1250_Hat's Fibonacci【大数】【水题】
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.
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.
Author
戴帽子的
#include<stdio.h> #include<string.h> int f1[2050],f2[2050],f3[2050],f4[2050],f5[2050]; int main() { int N; while(~scanf("%d",&N)) { if(N==1) printf("1\n"); else if(N==2) printf("1\n"); else if(N==3) printf("1\n"); else if(N==4) printf("1\n"); else { memset(f1,0,sizeof(f1)); memset(f2,0,sizeof(f2)); memset(f3,0,sizeof(f3)); memset(f4,0,sizeof(f4)); memset(f5,0,sizeof(f5)); f1[0] = f2[0] = f3[0] = f4[0] = 1; for(int i = 5; i<= N; i++) { for(int j = 0; j <= 2010; j++) { f5[j] = f1[j] + f2[j] + f3[j] + f4[j]; if(f5[j] >= 10) { f1[j+1] += (f5[j]/10); f5[j] = (f5[j]%10); } f1[j] = f2[j]; f2[j] = f3[j]; f3[j] = f4[j]; f4[j] = f5[j]; } } int j; for(j = 2010; j >= 0; j--) if(f5[j]!=0) break; for(int i = j; i >= 0; i--) printf("%d",f5[i]); printf("\n"); } } return 0; }
HDU1250_Hat's Fibonacci【大数】【水题】
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