提交时间:2025-03-29 16:25:40

运行 ID: 63520

#include<bits/stdc++.h> using namespace std; int main() { int fib[1000010] = {1,1}; int n,a; cin>>n; // 先计算出1-1000000所有斐波那契数 for(int i = 2; i <= 1000000; i ++) fib[i] = fib[i-1] + fib[i-2]; for(int i = 1; i <= n; i ++) { cin>>a; cout<<fib[a-1]%1000<<endl; } return 0; }