入门训练 Fibonacci数列

PS:避免结果溢出,在每次计算过程中就对10007取余.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
    
int main() {
	ll n;
	cin>>n;
	ll b[n+1];
	b[1]=b[2]=1;

	for(int i=3; i<=n; i++){
		b[i]=(b[i-1]+b[i-2])%10007;	
	}
	cout<<b[n]; 
	return 0;
}

标签：PS,10007,入门,int,ll,long,Fibonacci,取余,数列
来源： https://blog.csdn.net/weixin_42306122/article/details/104702008