题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6356
Glad You Game
Steve has an integer array
Let the (j=li,li+1,⋯,ri), where
Let the (j=li,li+1,⋯,ri), where
{li=min((f3i−2modn)+1,(f3i−1modn)+1)ri=max((f3i−2modn)+1,(f3i−1modn)+1)vi=f3imod230(i=1,2,⋯,m).
InputThe first line contains one integer 5⋅107.OutputFor each test case, output the answer in one line.
Sample Input
4 1 10 100 1000 10000 10 100 1000 10000 100000 100 1000 10000 100000 1000000 1000 10000 100000 1000000 10000000
Sample Output
1031463378
1446334207
351511856
47320301347
Hint
In the first sample, a = [1031463378] after all the operations.
In the second sample, a = [1036205629, 1064909195, 1044643689, 1062944339, 1062944339, 1062944339, 1062944339, 1057472915, 1057472915, 1030626924] after all the operations.
题意：给你一个生成L,R,V 的函数，让你对区间L~R的数，把小于v的数都变为v,最后求所有数的异或；
题解：我们利用线段树维护每一段区间的最大值，最小值，对于一段区间L~R，如果最大值都小于V，那么久更新整个区间，如果最小值都大于V，就不操作；最后暴力查找每个位置的数异或求值即可；
参考代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn=1e5+10; 5 int n,m,T; 6 unsigned int X,Y,Z; 7 8 struct Node{ 9 int l,r,maxnum,minnum,tag; 10 } tree[maxn<<2]; 11 12 unsigned int functions() 13 { 14 X=X^(X<<11); 15 X=X^(X>>4); 16 X=X^(X<<5); 17 X=X^(X>>14); 18 unsigned int w=X^(Y^Z); 19 X=Y; Y=Z; Z=w; 20 return Z; 21 } 22 23 void build(int pos,int l,int r) 24 { 25 tree[pos].l=l,tree[pos].r=r,tree[pos].tag=-1; 26 if(l==r) 27 { 28 tree[pos].maxnum=0; 29 tree[pos].minnum=0; 30 return ; 31 } 32 int mid=(tree[pos].l+tree[pos].r)>>1; 33 build(pos<<1,l,mid); 34 build(pos<<1|1,mid+1,r); 35 tree[pos].maxnum=max(tree[pos<<1].maxnum,tree[pos<<1|1].maxnum); 36 tree[pos].minnum=min(tree[pos<<1].minnum,tree[pos<<1|1].minnum); 37 } 38 39 void pushup(int pos) 40 { 41 tree[pos].maxnum=max(tree[pos<<1].maxnum,tree[pos<<1|1].maxnum); 42 tree[pos].minnum=min(tree[pos<<1].minnum,tree[pos<<1|1].minnum); 43 } 44 45 void pushdown(int pos) 46 { 47 tree[pos<<1].maxnum=max(tree[pos<<1].maxnum,tree[pos].tag); 48 tree[pos<<1|1].maxnum=max(tree[pos<<1|1].maxnum,tree[pos].tag); 49 tree[pos<<1].minnum=max(tree[pos<<1].minnum,tree[pos].tag); 50 tree[pos<<1|1].minnum=max(tree[pos<<1|1].minnum,tree[pos].tag); 51 tree[pos<<1].tag=max(tree[pos<<1].tag,tree[pos].tag); 52 tree[pos<<1|1].tag=max(tree[pos<<1|1].tag,tree[pos].tag); 53 tree[pos].tag=-1; 54 } 55 56 void update(int pos,int l,int r,int val) 57 { 58 if(tree[pos].l==tree[pos].r) 59 { 60 tree[pos].maxnum=max(tree[pos].maxnum,val); 61 tree[pos].minnum=max(tree[pos].minnum,val); 62 return ; 63 } 64 if(tree[pos].l>=l&&tree[pos].r<=r) 65 { 66 if(tree[pos].maxnum<=val) 67 { 68 tree[pos].maxnum=tree[pos].minnum=val; 69 tree[pos].tag=max(tree[pos].tag,val); 70 return ; 71 } 72 } 73 if(tree[pos].minnum>=val) return ; 74 75 if(tree[pos].tag!=-1) pushdown(pos); 76 int mid=(tree[pos].l+tree[pos].r)>>1; 77 if(r<=mid) update(pos<<1,l,r,val); 78 else if(l>=mid+1) update(pos<<1|1,l,r,val); 79 else update(pos<<1,l,mid,val),update(pos<<1|1,mid+1,r,val); 80 pushup(pos); 81 } 82 83 int query(int pos,int k) 84 { 85 if(tree[pos].l==tree[pos].r&&tree[pos].l==k) return tree[pos].maxnum; 86 int mid=(tree[pos].l+tree[pos].r)>>1; 87 if(tree[pos].tag!=-1) pushdown(pos); 88 if(k<=mid) return query(pos<<1,k); 89 else return query(pos<<1|1,k); 90 } 91 92 int main() 93 { 94 ios::sync_with_stdio(false); 95 cin.tie(0); 96 cin>>T; 97 while(T--) 98 { 99 LL ans=0; 100 cin>>n>>m>>X>>Y>>Z; 101 build(1,1,n); 102 for(int i=1;i<=m;i++) 103 { 104 int l=functions()%n+1; 105 int r=functions()%n+1; 106 int v=functions()%(1<<30); 107 if(l>r) swap(l,r); 108 update(1,l,r,v); 109 } 110 for(int i=1;i<=n;i++) ans^=1ll*i*query(1,i); 111 cout<<ans<<endl; 112 } 113 114 return 0; 115 }