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。~
1 #include <bits/stdc++.h> 2 #define pi acos(-1) 3 #define fastcin ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL); 4 using namespace std; 5 typedef long long LL; 6 typedef pair<int, int> PII; 7 const int INF = 0x3f3f3f3f; // 不能加负号！！！ 8 const LL LL_INF = 0x3f3f3f3f3f3f3f3f;//4e18 ~= 2^62 9 const int maxn =100000 + 10; 10 const LL mod = pow(2, 30); 11 LL maxx[maxn<<2]; 12 LL a[maxn]; 13 LL ans ; 14 unsigned int RNG61(unsigned int &x, unsigned int &y, unsigned int &z) 15 { 16 x = x^(x<<11); 17 x = x^(x>>4); 18 x = x^(x<<5); 19 x = x^(x>>14); 20 unsigned int w = x^(y^z); 21 x = y; 22 y = z; 23 z = w; 24 return z; 25 } 26 27 void PushDown(int rt) 28 { 29 maxx[rt<<1] = max(maxx[rt<<1],maxx[rt]); 30 maxx[rt<<1|1] = max(maxx[rt<<1|1],maxx[rt]); 31 } 32 33 //区间更新，假设A[L,R]+=C 34 void update(int L, int R, LL v, int l, int r, int rt) 35 { 36 //cout<<l<<"____"<<r<<"....."<<rt<<endl; 37 if(maxx[rt] > v) return; 38 if(L<=l && r<=R){ 39 maxx[rt] = max(maxx[rt],v); 40 return ; 41 } 42 int mid = (l+r)>>1; 43 PushDown(rt); 44 if(L <= mid) update(L, R, v, l, mid, rt<<1); 45 if(R > mid) update(L, R, v, mid+1, r, rt<<1|1); 46 // PushUp(rt); 47 } 48 void get(int l,int r,int rt){ 49 if(l == r){ 50 ans ^= l*maxx[rt]; 51 return ; 52 } 53 PushDown(rt); 54 int mid = (l+r)>>1; 55 get(l,mid,rt<<1); 56 get(mid+1,r,rt<<1|1); 57 } 58 int main() 59 { 60 int T; 61 cin>>T; 62 while(T--){ 63 int n, m, l, r, v; 64 memset(maxx,0,sizeof(maxx)); 65 memset(a,0,sizeof(a)); 66 unsigned int X, Y, Z; 67 scanf("%d%d%u%u%u", &n, &m, &X, &Y, &Z); 68 unsigned int t1, t2, t3; 69 for(int i=0; i<m; i++){ 70 t1 = RNG61(X, Y, Z); 71 t2 = RNG61(X, Y, Z); 72 t3 = RNG61(X, Y, Z); 73 l = min((t1%n)+1, (t2%n)+1); 74 r = max((t1%n)+1, (t2%n)+1); 75 v = (t3 % mod + mod) %mod; 76 update(l, r, v, 1, n, 1); 77 } 78 ans = 0; 79 get(1,n,1); 80 printf("%lld ", ans); 81 } 82 }