1214. 波动数列

Posted 2021-09-06 幽殇默

https://www.acwing.com/solution/content/9223/

https://www.acwing.com/solution/content/7507/

目标状态：合法转移到s，共有n−1步，故最终答案为f[n−1][s%n]

#include<bits/stdc++.h>
using namespace std;
const int N=1010,mod=1e8+7;
int n,s,a,b,f[N][N];
int get_mod(int a,int b)
{
	return (a%b+b)%b;
}
int main(void)
{
	cin>>n>>s>>a>>b;
	f[0][0]=1;
	for(int i=1;i<n;i++)
	{
		for(int j=0;j<n;j++)
		{
			f[i][j]=(f[i-1][get_mod(j-(n-i)*a,n)]+f[i-1][get_mod(j+(n-i)*b,n)])%mod;
		}
	}
	cout<<f[n-1][get_mod(s,n)];
	return 0;
}