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hdu 4578 Transformation(线段树)
Transformation
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 3084 Accepted Submission(s): 749
Problem Description
Yuanfang is puzzled with the question below:
There are n integers, a1, a2, …, an. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between ax and ay inclusive. In other words, do transformation ak<---ak+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation ak<---ak×c, k = x,x+1,…,y.
Operation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation ak<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of axp+ax+1p+…+ay p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.
There are n integers, a1, a2, …, an. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between ax and ay inclusive. In other words, do transformation ak<---ak+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation ak<---ak×c, k = x,x+1,…,y.
Operation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation ak<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of axp+ax+1p+…+ay 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.
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
写得比较恶心。。因为 p <= 3 ， 所以处理好3个sum值 。
push_down要写得优美才可以过
至于，过程中那些 + 与 * 的操作的话只是需要把 n 方公式分解好就可以了。
比如 说 （ c * a + b ）^ 3 = (c*a)^3 + 3*(c*a)^2*b + 3*(a*c)*b^2 + b^3 .
那么平方 ， 一次的操作也是这么进行
至于操作3的话就是直接把 操作1跟操作2的lazy清空掉就可以了
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstring>#include <map>#include <queue>using namespace std;typedef long long LL ;typedef pair<int,int> pii ;#define X first#define Y second#define root 1,n,1#define lr rt<<1#define rr rt<<1|1#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1const int N = 200010;const int mod = 10007;int sum1[N<<2] , sum2[N<<2] , sum3[N<<2] , lazy1[N<<2] , lazy2[N<<2] , lazy3[N<<2];int n , m ;void build( int l , int r , int rt ) { sum1[rt] = sum2[rt] = sum3[rt] = 0 ; lazy1[rt] = lazy3[rt] = 0 ; // add and clean lazy2[rt] = 1 ; // muti if( l == r ) return ; int mid = (l+r)>>1; build(lson) , build(rson);}void Up( int rt ) { sum1[rt] = ( sum1[lr] + sum1[rr] ) % mod ; sum2[rt] = ( sum2[lr] + sum2[rr] ) % mod ; sum3[rt] = ( sum3[lr] + sum3[rr] ) % mod ;}void Down( int l , int r , int rt ) { if( l == r ) return ; int mid = (l+r)>>1; if( lazy3[rt] != 0 ) { lazy3[lr] = lazy3[rr] = lazy3[rt] ; lazy1[lr] = lazy1[rr] = 0 ; lazy2[lr] = lazy2[rr] = 1 ; sum1[lr] = ( mid - l + 1 ) * lazy3[rt] % mod ; sum2[lr] = ( mid - l + 1 ) * lazy3[rt] % mod * lazy3[rt] % mod ; sum3[lr] = ( mid - l + 1 ) * lazy3[rt] % mod * lazy3[rt] % mod * lazy3[rt] % mod ; sum1[rr] = ( r - mid ) * lazy3[rt] % mod ; sum2[rr] = ( r - mid ) * lazy3[rt] % mod * lazy3[rt] % mod ; sum3[rr] = ( r - mid ) * lazy3[rt] % mod * lazy3[rt] % mod * lazy3[rt] % mod ; lazy3[rt] = 0 ; } if( lazy1[rt] != 0 || lazy2[rt] != 1 ) { lazy1[lr] = ( lazy1[lr] * lazy2[rt] % mod + lazy1[rt] ) % mod ; lazy2[lr] = lazy2[lr] * lazy2[rt] % mod ; sum3[lr] = ( lazy2[rt] * lazy2[rt] % mod * lazy2[rt] % mod * sum3[lr] % mod + 3 * lazy2[rt] % mod * lazy2[rt] % mod * sum2[lr] % mod * lazy1[rt] % mod + + 3 * lazy2[rt] % mod * sum1[lr] % mod * lazy1[rt] % mod * lazy1[rt] % mod + ( mid - l + 1 ) * lazy1[rt] % mod * lazy1[rt] % mod * lazy1[rt] % mod ) % mod ; sum2[lr] = ( lazy2[rt] * lazy2[rt] % mod * sum2[lr] % mod + 2 * lazy1[rt] % mod * lazy2[rt] % mod * sum1[lr] % mod + ( mid - l + 1 ) * lazy1[rt] % mod * lazy1[rt] % mod ) % mod ; sum1[lr] = ( sum1[lr] * lazy2[rt] % mod + lazy1[rt]*( mid - l + 1 ) % mod ) % mod; lazy1[rr] = ( lazy1[rr] * lazy2[rt] % mod + lazy1[rt] ) % mod ; lazy2[rr] = lazy2[rr] * lazy2[rt] % mod ; sum3[rr] = ( lazy2[rt] * lazy2[rt] % mod * lazy2[rt] % mod * sum3[rr] % mod + 3 * lazy2[rt] % mod * lazy2[rt] % mod * sum2[rr] % mod * lazy1[rt] % mod + + 3 * lazy2[rt] % mod * sum1[rr] % mod * lazy1[rt] % mod * lazy1[rt] % mod + ( r - mid ) * lazy1[rt] % mod * lazy1[rt] % mod * lazy1[rt] % mod ) % mod ; sum2[rr] = ( lazy2[rt] * lazy2[rt] % mod * sum2[rr] % mod + 2 * lazy1[rt] % mod * lazy2[rt] % mod * sum1[rr] % mod + ( r - mid ) * lazy1[rt] % mod * lazy1[rt] % mod ) % mod ; sum1[rr] = ( sum1[rr] * lazy2[rt] % mod + lazy1[rt]*( r - mid ) % mod ) % mod; lazy1[rt] = 0; lazy2[rt] = 1; }}void update( int l , int r , int rt , int L , int R , int c , int op ) { if( l == L && r == R ) { // suppose lazy1 = lazy2 = 0 ; op1 c %= mod ; if( op == 1 ) { lazy1[rt] = ( c + lazy1[rt] ) % mod; sum3[rt] = ( sum3[rt] + 3 * sum2[rt] % mod * c % mod + 3 * sum1[rt] % mod * c % mod * c % mod + c * c % mod * c % mod * ( r - l + 1 ) % mod ) % mod; sum2[rt] = ( sum2[rt] + 2 * sum1[rt] % mod * c % mod + c * c % mod * ( r - l + 1 ) % mod ) % mod ; sum1[rt] = ( sum1[rt] + ( r - l + 1 ) * c % mod ) % mod; } else if( op == 2 ) { lazy1[rt] = lazy1[rt] * c % mod ; lazy2[rt] = lazy2[rt] * c % mod ; sum1[rt] = sum1[rt] * c % mod ; sum2[rt] = sum2[rt] * c % mod * c % mod ; sum3[rt] = sum3[rt] * c % mod * c % mod * c % mod ; } else { lazy1[rt] = 0 ; lazy2[rt] = 1 ; lazy3[rt] = c % mod ; sum1[rt] = (r-l+1) * c % mod ; sum2[rt] = (r-l+1) * c % mod * c % mod; sum3[rt] = (r-l+1) * c % mod * c % mod *c % mod; } return ; } Down(l,r,rt); int mid = (l+r)>>1; if( R <= mid ) update(lson,L,R,c,op); else if( L > mid ) update(rson,L,R,c,op); else update(lson,L,mid,c,op),update(rson,mid+1,R,c,op); Up(rt);}int query( int l , int r , int rt , int L , int R , int c ) { if( l == L && r == R ) { if( c == 1 ) return sum1[rt] ; else if( c == 2 ) return sum2[rt]; else return sum3[rt]; } Down(l,r,rt); int mid = (l+r)>>1; if( R <= mid ) return query(lson,L,R,c); else if( L > mid ) return query(rson,L,R,c); else return (query(lson,L,mid,c)+query(rson,mid+1,R,c))%mod;}int main(){ #ifdef LOCAL freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); #endif // LOCAL while( ~scanf("%d%d",&n,&m ) ) { if( n == 0 && m == 0 ) break ; build( root ) ; int op , x , y , c ; while( m-- ) { scanf("%d%d%d%d",&op,&x,&y,&c); if( op != 4 ) update(root,x,y,c,op); else printf("%d\n",query(root,x,y,c)); } }}
hdu 4578 Transformation(线段树)
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