hdu 5737

Differencia

Time Limit: 10000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 997    Accepted Submission(s): 293

 

Problem Description

Professor Zhang has two sequences a1,a2,...,an and b1,b2,...,bn. He wants to perform two kinds of operations on the sequences:

1. + l r x: set ai to x for all l≤i≤r.
2. ? l r: find the number of i such that ai≥bi and l≤i≤r.

 

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains four integers n, m, A and B (1≤n≤105,1≤m≤3000000,1≤A,B≤216) -- the length of the sequence, the number of operations and two parameters.

The second line contains n integers a1,a2,...,an (1≤ai≤109). The third line contains n integers b1,b2,...,bn (1≤bi≤109). 

As there are too many operations, the m operations are specified by parameters A and B given to the following generator routine.

int a = A, b = B, C = ~(1<<31), M = (1<<16)-1;
int rnd(int last) {
    a = (36969 + (last >> 3)) * (a & M) + (a >> 16);
    b = (18000 + (last >> 3)) * (b & M) + (b >> 16);
    return (C & ((a << 16) + b)) % 1000000000;
}

For the i-th operation, first call rnd(last) three times to get l, r and x (i.e. l = rnd(last) % n + 1, r = rnd(last) % n + 1, x = rnd(last) + 1). Then if l>r, you should swap their value. And at last, the i-th operation is type ?, if (l+r+x) is an even number, or type + otherwise.

Note: last is the answer of the latest type ? operation and assume last=0 at the beginning of each test case.

 

 

Output

For each test case, output an integer S=(∑i=1mi⋅zi) mod (109+7), where zi is the answer for i-the query. If the i-th query is of type +, then zi=0.

 

 

Sample Input

3 5 10 1 2 5 4 3 2 1 1 2 3 4 5 5 10 3 4 5 4 4 2 1 1 2 3 4 5 5 10 5 6 5 4 5 2 1 1 2 2 4 5

 

 

Sample Output

81 88 87

 

 

Author

zimpha

 

 

Source

2016 Multi-University Training Contest 2

 

題意：

給了A和B數組，有兩個操作，第一個操作是將A數組從l到r區間所有數置爲x，第二個操作是詢問A數組從l到r區間有多少個a[i]>=b[i]的。有m次操作，每次操作都是根據題目給的函數生成出來的。

 

思路：

對這道題我是沒什麼思路的，看了其他人的博客，才明白線段樹套有序表這麼巧妙的組合。在創建線段樹的過程中，有序數組就可以創建出來。並且創造出父親與左右兒子之間的映射關係，有點歸併排序的意思。

線段樹中創建的有序表是存的B數組，對於第一個操作，我們只需要把sum數組進行更新就行了，每次往下找的時候都是根據之前預處理好的對應關係來向下找到對應的位置，就可以直接更新sum數組。第二個操作就是基本的線段樹操作。

 

代碼：

#include <bits/stdc++.h>
#define ls i<<1
#define rs ls|1
#define mid (l + r >> 1)
#define lson i<<1,l,mid
#define rson i<<1|1,mid+1,r
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
int a1[maxn],a2[maxn];
int san[maxn*20],st[maxn*4],ed[maxn*4],sum[maxn*4],down[maxn*4];
int to[2][maxn*20];
int a,b,C,M;
int cnt;
void pushup(int i){sum[i]=sum[ls]+sum[rs];}
void pushdown(int i)
{
    if(~down[i])
    {
        down[ls]=to[0][down[i]];
        sum[ls]=down[ls]-st[ls]+1;

        down[rs]=to[1][down[i]];
        sum[rs]=down[rs]-st[rs]+1;
        down[i]=-1;
    }
}
void build(int i,int l,int r)
{
    down[i]=-1;
    san[cnt++]=-inf;
    st[i]=cnt;
    for(int i=l;i<=r;i++) san[cnt++]=a2[i];
    ed[i]=cnt;
    sort(san+st[i],san+ed[i]);
    if(l==r)
    {
        sum[i]=(a1[l]>=a2[l]);
        return;
    }
    build(lson);
    build(rson);

    int lx=st[ls],rx=st[rs];
    for(int j=st[i];j<ed[i];j++)
    {
        while(lx<ed[ls]&&san[lx]<=san[j]) lx++;
        while(rx<ed[rs]&&san[rx]<=san[j]) rx++;
        to[0][j]=lx-1;
        to[1][j]=rx-1;
    }
    to[0][st[i]-1]=st[ls]-1;
    to[1][st[i]-1]=st[rs]-1;
    pushup(i);
}
void update(int i,int l,int r,int x,int y,int pos)
{
    if(x<=l&&r<=y)
    {
        sum[i]=pos-st[i]+1;
        down[i]=pos;
        return;
    }
    pushdown(i);
    if(x<=mid) update(lson,x,y,to[0][pos]);
    if(y>mid) update(rson,x,y,to[1][pos]);
    pushup(i);
}
int query(int i,int l,int r,int x,int y)
{
    int ans=0;
    if(x<=l&&r<=y) return sum[i];
    pushdown(i);
    if(x<=mid) ans+=query(lson,x,y);
    if(y>mid) ans+=query(rson,x,y);
    pushup(i);
    return ans;
}
int rnd(int last)
{
    a = (36969 + (last >> 3)) * (a & M) + (a >> 16);
    b = (18000 + (last >> 3)) * (b & M) + (b >> 16);
    return (C & ((a << 16) + b)) % 1000000000;
}
int main()
{
    int t,n,m,A,B,x;
    scanf("%d",&t);
    while(t--)
    {
        cnt=0;
        int last=0,ans=0;
        scanf("%d%d%d%d",&n,&m,&a,&b);
        for(int i=1;i<=n;i++) scanf("%d",&a1[i]);
        for(int i=1;i<=n;i++) scanf("%d",&a2[i]);
        build(1,1,n);
        for(int i=1;i<=m;i++)
        {
            C = ~(1<<31),M = (1<<16)-1;
            int l=rnd(last)%n+1,r=rnd(last)%n+1,x=rnd(last)+1;
            if(l>r) swap(l,r);
            if((l+r+x)%2==0)// ?
            {
                last=query(1,1,n,l,r);
                int tmp=((ll)last*i)%mod;
                ans=(ans+tmp)%mod;
            }
            else
            {
                x=upper_bound(san+st[1],san+ed[1],x)-san;
                update(1,1,n,l,r,x-1);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}