杭電ACM-1021 Fibonacci Again
阿新 • • 發佈：2018-11-26
Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 75703 Accepted Submission(s): 34365
Problem Description
There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).
Input
Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).
Output
Print the word "yes" if 3 divide evenly into F(n).
Print the word "no" if not.
Sample Input
0
1
2
3
4
5
Sample Output
no
no
yes
no
no
no
斐波那契數列擴充套件，還是一個水題，題不難，關鍵在於10^6會溢位，long long 也過不了。
題目要求判斷F [ i ] % 3 是否為 0,數列已給出
F(0) = 7
F(1) = 11
F(n) = F(n-1) + F(n-2) (n>=2)
既然數字太多過不了，那麼我們考慮找規律，這裡有一個規律是：
兩個數的餘數之和的餘數等於兩個數和的餘數
即：（ f [ i-1 ]%3+f [ i-2 ]%3 ）%3 = f [ i ] %3
AC程式碼奉上：
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
long long int f[1000001];
void fun()
{
int i;
f[0] = 7;f[1] = 11;
f[0] % = 3;f[1] % = 3;
for(i = 2;i <= 1000000;i++)
{
f[i] = f[i-1] + f[i-2];
f[i] % = 3;
}
}
int main()
{
int n;
int a,b;
fun();
while(cin>>n)
{
if(f[n] == 0)
{
cout<<"yes"<<endl;
}
else
cout<<"no"<<endl;
}
return 0;
}