Glad You Came hdu—6356（线段树维护最小值）

Glad You Came

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 768    Accepted Submission(s): 264

Problem Description

Steve has an integer array a of length n (1-based). He assigned all the elements as zero at the beginning. After that, he made m operations, each of which is to update an interval of a with some value. You need to figure out ⨁ni=1(i⋅ai) after all his operations are finished, where ⨁ means the bitwise exclusive-OR operator.
In order to avoid huge input data, these operations are encrypted through some particular approach.
There are three unsigned 32-bit integers X,Y and Z which have initial values given by the input. A random number generator function is described as following, where ∧ means the bitwise exclusive-OR operator, << means the bitwise left shift operator and >> means the bitwise right shift operator. Note that function would change the values of X,Y and Z after calling.

Let the i-th result value of calling the above function as fi (i=1,2,⋯,3m). The i-th operation of Steve is to update aj as vi if aj<vi (j=li,li+1,⋯,ri), where

⎧⎩⎨⎪⎪lirivi=min((f3i−2modn)+1,(f3i−1modn)+1)=max((f3i−2modn)+1,(f3i−1modn)+1)=f3imod230(i=1,2,⋯,m).

Input

The first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains five space-separated integers n,m,X,Y and Z.
1≤T≤100, 1≤n≤105, 1≤m≤5⋅106, 0≤X,Y,Z<230.
It is guaranteed that the sum of n in all the test cases does not exceed 106 and the sum of m in all the test cases does not exceed 5⋅107.

Output

For 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.

 

 

Source

2018 Multi-University Training Contest 5

 

 

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这道题是简单。。题。

跟之前思路一样，维护一个最小值，当最小值都比v小，就不更新了

反之一直更新，直到叶子节点。查询的时候单点查询。

没想到这个随机数来得巧，这么暴力都能过。看来是修改操作是很小的。

代码：2496MS

#include 
#define ll long long
#define lson l,m,id<<1
#define rson m+1,r,id<<1|1
using namespace std;
const int maxn=1e5+10;
const int INF=0x3f3f3f3f;
const int mod=(1<<30);
unsigned  x,y,z;
unsigned  f[15000010];
ll minn[maxn<<2];
unsigned  get()
{
    x=x^(x<<11);
    x=x^(x>>4);
    x=x^(x<<5);
    x=x^(x>>14);
    unsigned w=x^(y^z);
    x=y;
    y=z;
    z=w;
    return z;
}
void pushup(int id)
{
    minn[id]=min(minn[id<<1],minn[id<<1|1]);
}
void update(int L,int R,ll val,int l,int r,int id){
    if(minn[id]>=val)
        return;
    if(l==r){
        minn[id]=val;
        return ;
    }
    int m=(l+r)/2;
    if(m>=L)
        update(L,R,val,lson);
    if(m

使用lazy标记，就是在维护一个最大值。当最大值都小于v，说明里面的所有的值都要更新成v

大佬的代码：反而慢了。。。。。

代码：3182MS

#include 
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn=1e5+10;
const int INF=0x3f3f3f3f;
const int mod=(1<<30);
unsigned  x,y,z;
unsigned  f[15000010];

ll max_1[maxn<<2];
ll la[maxn<<2];

void pushup(int rt)
{
    max_1[rt]=max(max_1[rt<<1],max_1[rt<<1|1]);
}
void pushdown(int rt)
{
    if(la[rt])
    {
        max_1[rt<<1]=max_1[rt<<1|1]=la[rt];
        la[rt<<1]=la[rt<<1|1]=la[rt];
        la[rt]=0;
    }
}

void update(int L,int R,ll v,int l,int r,int rt)
{
    if(l==r&&max_1[rt]>v)return;
    if(L<=l&&r<=R&&max_1[rt]<=v)
    {
        max_1[rt]=v;
        la[rt]=v;
        return;
    }
    pushdown(rt);
    int m=(l+r)/2;
    if(m>=L)update(L,R,v,lson);
    if(m>4);
    x=x^(x<<5);
    x=x^(x>>14);
    unsigned w=x^(y^z);
    x=y;
    y=z;
    z=w;
    return z;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        memset(max_1,0,sizeof(max_1));
        memset(la,0,sizeof(la));
        scanf("%d%d%d%d%d",&n,&m,&x,&y,&z);
        for(int i=1;i<=3*m;i++)
        {
            f[i]=get();
        }
        for(int i=1;i<=m;i++)
        {
            int l=min(f[3*i-2]%n+1,f[3*i-1]%n+1);
            int r=max(f[3*i-2]%n+1,f[3*i-1]%n+1);
            int w=f[3*i]%mod;
            update(l,r,w,1,n,1);
        }
        ll sum=0;
        for(int i=1;i<=n;i++)
        {
            ll w=getsum(i,i,1,n,1);
            //cout<