HDU 4578 线段树的四种操作

Transformation

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 3790    Accepted Submission(s): 924

Problem Description

Yuanfang is puzzled with the question below:
There are n integers, a ₁, a ₂, …, a _n. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between a _x and a _y inclusive. In other words, do transformation a _k<---a _k+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between a _x and a _y inclusive. In other words, do transformation a _k<---a _k×c, k = x,x+1,…,y.
Operation 3: Change the numbers between a _x and a _y to c, inclusive. In other words, do transformation a _k<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between a _x and a _y inclusive. In other words, get the result of a _x ^p+a _x+1 ^p+…+a _y ^p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.

 

Input

There are no more than 10 test cases.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.

 

Output

For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.

 

Sample Input

5 5 3 3 5 7 1 2 4 4 4 1 5 2 2 2 5 8 4 3 5 3 0 0

 

Sample Output

307 7489

 

Source

2013ACM-ICPC杭州赛区全国邀请赛

 

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四种操作。

1.[L,R]范围内的数全部+c
2.[L,R]范围内的数全部*c
3.[L,R]范围内的数全部变为c
4.输出[L,R]范围内的数的k次方和（1 <= k <= 3 )

#include 
#include 
#include 
using namespace std ;

#define lson l , m , o << 1
#define rson m + 1 , r , o << 1 | 1

const int mod = 10007 ;
const int maxN = 500000 ;

int sum1[maxN] , sum2[maxN] , sum3[maxN] ;
int add[maxN] , mul[maxN] , set[maxN] ;

void PushUp ( int o )
{
    sum1[o] = ( sum1[o << 1] + sum1[o << 1 | 1] ) % mod ;
    sum2[o] = ( sum2[o << 1] + sum2[o << 1 | 1] ) % mod ;
    sum3[o] = ( sum3[o << 1] + sum3[o << 1 | 1] ) % mod ;
}

void PushDown ( int o , int l , int r )
{
    int m = ( l + r ) >> 1 ;
    if ( set[o] )
    {
        set[o << 1] = set[o << 1 | 1] = set[o] ;
        add[o << 1] = add[o << 1 | 1] = 0 ;
        mul[o << 1] = mul[o << 1 | 1] = 1 ;
        sum1[o << 1] = ( m - l + 1 ) * set[o] % mod ;
        sum2[o << 1] = sum1[o << 1] * set[o] % mod ;
        sum3[o << 1] = sum2[o << 1] * set[o] % mod ;
        sum1[o << 1 | 1] = ( r - m ) * set[o] % mod ;
        sum2[o << 1 | 1] = sum1[o << 1 | 1] * set[o] % mod ;
        sum3[o << 1 | 1] = sum2[o << 1 | 1] * set[o] % mod ;
        set[o] = 0 ;
    }
    if ( add[o] || mul[o] != 1 )
    {
        add[o << 1] = ( add[o << 1] * mul[o] + add[o] ) % mod ;
        mul[o << 1] = mul[o << 1] * mul[o] % mod ;
        add[o << 1 | 1] = ( add[o << 1 | 1] * mul[o] + add[o] ) % mod ;
        mul[o << 1 | 1] = mul[o << 1 | 1] * mul[o] % mod ;

        int _sum1 , _sum2 , _sum3 ;

        _sum1 = ( sum1[o << 1] * mul[o] % mod + add[o] * ( m - l + 1 ) % mod ) % mod ;
        _sum2 = ( mul[o] * mul[o] % mod * sum2[o << 1] % mod
                  + 2 * add[o] % mod * mul[o] % mod * sum1[o << 1] % mod
                  + add[o] * add[o] % mod * ( m - l + 1 ) % mod ) % mod ;
        _sum3 = ( mul[o] * mul[o] % mod * mul[o] % mod * sum3[o << 1] % mod
                  + 3 * mul[o] % mod * mul[o] % mod * add[o] % mod * sum2[o << 1] % mod
                  + 3 * mul[o] % mod * add[o] % mod * add[o] % mod * sum1[o << 1] % mod
                  + add[o] * add[o] % mod * add[o] % mod * ( m - l + 1 ) % mod ) % mod ;

        sum1[o << 1] = _sum1 ;
        sum2[o << 1] = _sum2 ;
        sum3[o << 1] = _sum3 ;

        _sum1 = ( sum1[o << 1 | 1] * mul[o] % mod + add[o] * ( r - m ) % mod ) % mod ;
        _sum2 = ( mul[o] * mul[o] % mod * sum2[o << 1 | 1] % mod
                  + 2 * add[o] % mod * mul[o] % mod * sum1[o << 1 | 1] % mod
                  + add[o] * add[o] % mod * ( r - m ) % mod ) % mod ;
        _sum3 = ( mul[o] * mul[o] % mod * mul[o] % mod * sum3[o << 1 | 1] % mod
                  + 3 * mul[o] % mod * mul[o] % mod * add[o] % mod * sum2[o << 1 | 1] % mod
                  + 3 * mul[o] % mod * add[o] % mod * add[o] % mod * sum1[o << 1 | 1] % mod
                  + add[o] * add[o] % mod * add[o] % mod * ( r - m ) % mod ) % mod ;

        sum1[o << 1 | 1] = _sum1 ;
        sum2[o << 1 | 1] = _sum2 ;
        sum3[o << 1 | 1] = _sum3 ;

        add[o] = 0 ;
        mul[o] = 1 ;
    }
}

void Build ( int l , int r , int o )
{
    sum1[o] = sum2[o] = sum3[o] = 0 ;
    add[o] = set[o] = 0 ;
    mul[o] = 1 ;
    if ( l == r ) return ;
    int m = ( l + r ) >> 1 ;
    Build ( lson ) ;
    Build ( rson ) ;
}

void Update ( int ch , int v , int L , int R , int l , int r , int o )
{
    if ( L <= l && r <= R )
    {
        int len = r - l + 1 ;
        if ( 1 == ch )
        {
            add[o] = ( add[o] + v ) % mod ;
            int _sum1 , _sum2 , _sum3 ;
            _sum1 = ( sum1[o] + v * len ) % mod ;
            _sum2 = ( sum2[o] + 2 * sum1[o] % mod * v % mod + v * v % mod * len % mod ) % mod ;
            _sum3 = ( sum3[o] + 3 * v % mod * v % mod * sum1[o] % mod
                      + 3 * v % mod * sum2[o] % mod + v * v % mod * v % mod * len % mod ) % mod ;
            sum1[o] = _sum1 ;
            sum2[o] = _sum2 ;
            sum3[o] = _sum3 ;
        }
        if ( 2 == ch )
        {
            mul[o] = mul[o] * v % mod ;
            add[o] = add[o] * v % mod ;
            sum1[o] = sum1[o] * v % mod ;
            sum2[o] = sum2[o] * v % mod * v % mod ;
            sum3[o] = sum3[o] * v % mod * v % mod * v % mod ;
        }
        if ( 3 == ch )
        {
            set[o] = v ;
            add[o] = 0 ;
            mul[o] = 1 ;
            sum1[o] = v * len % mod ;
            sum2[o] = v * sum1[o] % mod ;
            sum3[o] = v * sum2[o] % mod ;
        }
        return ;
    }
    PushDown ( o , l , r ) ;
    int m = ( l + r ) >> 1 ;
    if ( L <= m ) Update ( ch , v , L , R , lson ) ;
    if ( m <  R ) Update ( ch , v , L , R , rson ) ;
    PushUp ( o ) ;
}

int Query ( int v , int L , int R , int l , int r , int o )
{
    if ( L <= l && r <= R )
    {
        if ( 1 == v ) return sum1[o] ;
        if ( 2 == v ) return sum2[o] ;
        if ( 3 == v ) return sum3[o] ;
    }
    PushDown ( o , l , r ) ;
    int ans = 0 , m = ( l + r ) >> 1 ;
    if ( L <= m ) ans = ( ans + Query ( v , L , R , lson ) ) % mod ;
    if ( m <  R ) ans = ( ans + Query ( v , L , R , rson ) ) % mod ;
    return ans ;
}

void work ()
{
    int n , m , ch , v , l , r ;
    while ( ~scanf ( "%d%d" , &n , &m ) && ( n || m ) )
    {
        Build ( 1 , n , 1 ) ;
        while ( m -- )
        {
            scanf ( "%d%d%d%d" , &ch , &l , &r , &v ) ;
            if ( ch == 4 )
            {
                int ans = Query ( v , l , r , 1 , n , 1 ) ;
                printf ( "%d\n" , ans ) ;
            }
            else Update ( ch , v , l , r , 1 , n , 1 ) ;
        }
    }
}

int main ()
{
    work () ;
    return 0 ;
}