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; }