http://oj.ecustacm.cn/problem.php?id=1453&csrf=J1ZRnq0e4NfKIP03Vj6vdG7MeBdswHio
题解:
对每一项取10000的模就行了。
#include<iostream>
#include<cstdio>
#define ll long long
using namespace std;
int a[20190325]={
1,1,1};
int main(void)
{
ll sum=0;
int j;
for(j=3;j<20190324;j++)
{
a[j]=(a[j-1]+a[j-2]+a[j-3])%10000;
}
printf("%d",a[20190323]);
return 0;
}