HDU 1250 - Hat's Fibonacci(大数斐波那契)
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Hat's Fibonacci
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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
4203968145672990846840663646Note:No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.
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先预处理一下,每4位放在二维数组的一个位置,一排代表一个斐波那契数
注意最后一个位置先输出,否则最后一个位置也按4位输出不足4位的时候会WA
#include <iostream>#include <cstdio>#include <cstring>using namespace std;int f[8000][555];int main(){ memset(f,0,sizeof(f)); f[1][0]=f[2][0]=f[3][0]=f[4][0]=1; int temp=0; for(int i=5;i<8000;i++) { for(int j=0;j<555;j++) { temp+=f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j]; f[i][j]=temp%10000; temp/=10000; } } int n; while(~scanf("%d",&n)) { int i; for(i=554;i>=0;i--) { if(f[n][i]!=0) break; } printf("%d",f[n][i--]); for(;i>=0;i--) { printf("%.4d",f[n][i]); } printf("\n"); } return 0;}
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