hdu 5593 ZYB's Tree 树形dp ZYB's Tree

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)


Problem Description
i.
 
Input
In the first line there is the number of testcases 1000000
 
Output
For 100000 are only for two tests finally.
 
Sample Input
1 3 1 1 1
 
Sample Output
3
 
Source
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<bitset>
#include<time.h>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=5e5+10,M=1e6+10,inf=1e9+7,MOD=1e9+7;
const LL INF=1e18+10,mod=1e9+7;

vector<int>edge[N];
int n,a,b,k;;
int dp[N][12][2];
void dfs(int u)
{
    dp[u][0][0]=1;
    for(int i=0;i<edge[u].size();i++)
    {
        int v=edge[u][i];
        dfs(v);
        for(int j=1;j<=k;j++)
        dp[u][j][0]+=dp[v][j-1][0];
    }

}
void dfs2(int u)
{
    for(int i=0;i<edge[u].size();i++)
    {
        int v=edge[u][i];
        dp[v][0][1]=dp[v][0][0];
        for(int i=1;i<=k;i++)
        {
            if(i<2)dp[v][i][1]=dp[u][i-1][1]+dp[v][i][0];
            else dp[v][i][1]=dp[u][i-1][1]+dp[v][i][0]-dp[v][i-2][0];
        }
        dfs2(v);
    }
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d%d%d%d",&n,&k,&a,&b);
        for(int i=1;i<=n;i++)
            edge[i].clear();
        for(int i=2;i<=n;i++)
        {
            int x=(1LL*a*i+b)%(i-1)+1;
            edge[x].push_back(i);
        }
        dfs(1);
        for(int i=0;i<=k;i++)dp[1][i][1]=dp[1][i][0];
        dfs2(1);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            int sum=0;
            for(int j=0;j<=k;j++)
                sum+=dp[i][j][1];
            ans^=sum;
        }
        printf("%d
",ans);
    }
    return 0;
}