题目链接:
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
const int N = 1e6 + 10;
int f[N], pre[N];
int main()
{
ios::sync_with_stdio(false), cin.tie(nullptr);
int n;
cin >> n;
pre[0] = f[0] = 1;
for (int i = 1; i <= n; i++) {
f[i] = (i64)((i >= 1 ? pre[i - 1] : 0) + (i >= 3 ? pre[i - 3] : 0)) % 10000;
pre[i] = (i64)(pre[i - 1] + f[i]) % 10000;
}
cout << f[n];
return 0;
}
标签:pre,i64,10000,覆盖,墙壁,long,P1990,int
From: https://www.cnblogs.com/pangyou3s/p/18172763