HDU-6356 Glad You Came (线段树)

题目链接：Glad You Came

题意：数组有n个数初始为0，m个询问，每个询问给出L R V（按照给定函数生成），将数组的下标L到R的数与V取较大值，最后输出给定的公式结果。

题意：哇~打比赛的时候想用两个线段树去维护，一棵维护每个结点所代表区间的最大值，一棵维护每个结点所代表区间的异或和。不过这题比我想的要暴力~~~直接维护最大值，最后遍历整棵树将最大值pushdown下去就可以了。~

 #include <bits/stdc++.h>
 #define pi acos(-1)
 #define fastcin ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
 using namespace std;
 typedef long long LL;
 typedef pair<int, int> PII;
 const int INF = 0x3f3f3f3f;       // 不能加负号！！！
 const LL LL_INF = 0x3f3f3f3f3f3f3f3f;//4e18 ~= 2^62
 const int maxn = + ;
 const LL mod = pow(, );
 LL maxx[maxn<<];
 LL a[maxn];
 LL ans ;
 unsigned int RNG61(unsigned int &x, unsigned int &y, unsigned int &z)
 {
     x = x^(x<<);
     x = x^(x>>);
     x = x^(x<<);
     x = x^(x>>);
     unsigned int w = x^(y^z);
     x = y;
     y = z;
     z = w;
     return z;
 }

 void PushDown(int rt)
 {
     maxx[rt<<] = max(maxx[rt<<],maxx[rt]);
     maxx[rt<<|] = max(maxx[rt<<|],maxx[rt]);
 }

 //区间更新，假设A[L,R]+=C
 void update(int L, int R, LL v, int l, int r, int rt)
 {
     //cout<<l<<"____"<<r<<"....."<<rt<<endl;
     if(maxx[rt] > v) return;
     if(L<=l && r<=R){
         maxx[rt] = max(maxx[rt],v);
         return ;
     }
     int mid = (l+r)>>;
     PushDown(rt);
     if(L <= mid) update(L, R, v, l, mid, rt<<);
     if(R >  mid) update(L, R, v, mid+, r, rt<<|);
 //    PushUp(rt);
 }
 void get(int l,int r,int rt){
     if(l == r){
         ans ^= l*maxx[rt];
         return ;
     }
     PushDown(rt);
     int mid = (l+r)>>;
     get(l,mid,rt<<);
     get(mid+,r,rt<<|);
 }
 int main()
 {
     int T;
     cin>>T;
     while(T--){
         int n, m, l, r, v;
         memset(maxx,,sizeof(maxx));
         memset(a,,sizeof(a));
         unsigned int X, Y, Z;
         scanf("%d%d%u%u%u", &n, &m, &X, &Y, &Z);
         unsigned int t1, t2, t3;
         for(int i=; i<m; i++){
             t1 = RNG61(X, Y, Z);
             t2 = RNG61(X, Y, Z);
             t3 = RNG61(X, Y, Z);
             l = min((t1%n)+, (t2%n)+);
             r = max((t1%n)+, (t2%n)+);
             v = (t3 % mod + mod) %mod;
             update(l, r, v, , n, );
         }
         ans = ;
         get(,n,);
         printf("%lld\n", ans);
     }
 }