题解 | #了断局#
了断局
https://ac.nowcoder.com/acm/problem/15036
了断局
题目大意：
该数列为Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.
代码实现：
预处理前100项数存入long long数组里.
代码：
#include <bits/stdc++.h>
using namespace std;
const int N = 1e2+10;
long long s[N];
int main()
{
int n;
s[1] = 0;
s[2] = 1;
s[3] = 1;
for(int i = 4; i <= 100; i++) {
s[i] = s[i-1] + s[i-2] + s[i-3];
}
while(cin>>n) {
cout<<s[n]<<endl;
}
return 0;
}
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