线段树 I - Transformation 加乘优先级
I - Transformation
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.
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.
InputThere 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.
OutputFor 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 题目思路：
这个题目是一个裸的线段树，有四种操作，
第一种就是区间更新，在每一个数值+c
第二种就是每一个位置*c
第三种就是把每一个位置的数更新成c
第四种求每一个数的c次方的和
前面三种就是普通的线段树，最后一种因为c比较小，最大只有三所以就在结构体里面枚举三种情况就可以了。
第一种和第二张要设置两个lazy标志，第三种也要设置，但是如果第三种成立则之前的lazy标志都要删去，一共有了三种lazy标志。
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;
const int MOD = ;
const int MAXN = ;
struct Node
{
int l,r;
int sum1,sum2,sum3;
int lazy1,lazy2,lazy3;
}segTree[MAXN*];
void build(int i,int l,int r)
{
segTree[i].l = l;
segTree[i].r = r;
segTree[i].sum1 = segTree[i].sum2 = segTree[i].sum3 = ;
segTree[i].lazy1 = segTree[i].lazy3 = ;
segTree[i].lazy2 = ;
int mid = (l+r)/;
if(l == r)return;
build(i<<,l,mid);
build((i<<)|,mid+,r);
}
void push_up(int i)
{
if(segTree[i].l == segTree[i].r)
return;
segTree[i].sum1 = (segTree[i<<].sum1 + segTree[(i<<)|].sum1)%MOD;
segTree[i].sum2 = (segTree[i<<].sum2 + segTree[(i<<)|].sum2)%MOD;
segTree[i].sum3 = (segTree[i<<].sum3 + segTree[(i<<)|].sum3)%MOD; } void push_down(int i)
{
if(segTree[i].l == segTree[i].r) return;
if(segTree[i].lazy3 != )
{
segTree[i<<].lazy3 = segTree[(i<<)|].lazy3 = segTree[i].lazy3;
segTree[i<<].lazy1 = segTree[(i<<)|].lazy1 = ;
segTree[i<<].lazy2 = segTree[(i<<)|].lazy2 = ;
segTree[i<<].sum1 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD;
segTree[i<<].sum2 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD;
segTree[i<<].sum3 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD;
segTree[(i<<)|].sum1 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD;
segTree[(i<<)|].sum2 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD;
segTree[(i<<)|].sum3 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD;
segTree[i].lazy3 = ;
}
if(segTree[i].lazy1 != || segTree[i].lazy2 != )
{
segTree[i<<].lazy1 = ( segTree[i].lazy2*segTree[i<<].lazy1%MOD + segTree[i].lazy1 )%MOD;
segTree[i<<].lazy2 = segTree[i<<].lazy2*segTree[i].lazy2%MOD;
int sum1,sum2,sum3;
sum1 = (segTree[i<<].sum1*segTree[i].lazy2%MOD + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD)%MOD;
sum2 = (segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i<<].sum2 % MOD + *segTree[i].lazy1*segTree[i].lazy2%MOD * segTree[i<<].sum1%MOD + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD*segTree[i].lazy1%MOD)%MOD;
sum3 = segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i<<].sum3 % MOD;
sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i<<].sum2) % MOD;
sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD * segTree[i<<].sum1) % MOD;
sum3 = (sum3 + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD) % MOD;
segTree[i<<].sum1 = sum1;
segTree[i<<].sum2 = sum2;
segTree[i<<].sum3 = sum3;
segTree[(i<<)|].lazy1 = ( segTree[i].lazy2*segTree[(i<<)|].lazy1%MOD + segTree[i].lazy1 )%MOD;
segTree[(i<<)|].lazy2 = segTree[(i<<)|].lazy2 * segTree[i].lazy2 % MOD;
sum1 = (segTree[(i<<)|].sum1*segTree[i].lazy2%MOD + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD)%MOD;
sum2 = (segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[(i<<)|].sum2 % MOD + *segTree[i].lazy1*segTree[i].lazy2%MOD * segTree[(i<<)|].sum1%MOD + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD*segTree[i].lazy1%MOD)%MOD;
sum3 = segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[(i<<)|].sum3 % MOD;
sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[(i<<)|].sum2) % MOD;
sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD * segTree[(i<<)|].sum1) % MOD;
sum3 = (sum3 + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD) % MOD;
segTree[(i<<)|].sum1 = sum1;
segTree[(i<<)|].sum2 = sum2;
segTree[(i<<)|].sum3 = sum3;
segTree[i].lazy1 = ;
segTree[i].lazy2 = ; }
}
void update(int i,int l,int r,int type,int c)
{
if(segTree[i].l == l && segTree[i].r == r)
{
c %= MOD;
if(type == )
{
segTree[i].lazy1 += c;
segTree[i].lazy1 %= MOD;
segTree[i].sum3 = (segTree[i].sum3 + *segTree[i].sum2%MOD*c%MOD + *segTree[i].sum1%MOD*c%MOD*c%MOD + (segTree[i].r - segTree[i].l + )*c%MOD*c%MOD*c%MOD)%MOD;
segTree[i].sum2 = (segTree[i].sum2 + *segTree[i].sum1%MOD*c%MOD + (segTree[i].r - segTree[i].l + )*c%MOD*c%MOD)%MOD;
segTree[i].sum1 = (segTree[i].sum1 + (segTree[i].r - segTree[i].l + )*c%MOD)%MOD;
}
else if(type == )
{
segTree[i].lazy1 = segTree[i].lazy1*c%MOD;
segTree[i].lazy2 = segTree[i].lazy2*c%MOD;
segTree[i].sum1 = segTree[i].sum1*c%MOD;
segTree[i].sum2 = segTree[i].sum2*c%MOD*c%MOD;
segTree[i].sum3 = segTree[i].sum3*c%MOD*c%MOD*c%MOD;
}
else
{
segTree[i].lazy1 = ;
segTree[i].lazy2 = ;
segTree[i].lazy3 = c%MOD;
segTree[i].sum1 = c*(segTree[i].r - segTree[i].l + )%MOD;
segTree[i].sum2 = c*(segTree[i].r - segTree[i].l + )%MOD*c%MOD;
segTree[i].sum3 = c*(segTree[i].r - segTree[i].l + )%MOD*c%MOD*c%MOD;
}
return;
}
push_down(i);
int mid = (segTree[i].l + segTree[i].r)/;
if(r <= mid)update(i<<,l,r,type,c);
else if(l > mid)update((i<<)|,l,r,type,c);
else
{
update(i<<,l,mid,type,c);
update((i<<)|,mid+,r,type,c);
}
push_up(i);
}
int query(int i,int l,int r,int p)
{
if(segTree[i].l == l && segTree[i].r == r)
{
if(p == )return segTree[i].sum1;
else if(p== )return segTree[i].sum2;
else return segTree[i].sum3;
}
push_down(i);
int mid = (segTree[i].l + segTree[i].r )/;
if(r <= mid)return query(i<<,l,r,p);
else if(l > mid)return query((i<<)|,l,r,p);
else return (query(i<<,l,mid,p)+query((i<<)|,mid+,r,p))%MOD;
} int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n,m;
while(scanf("%d%d",&n,&m) == )
{
if(n == && m == )break;
build(,,n);
int type,x,y,c;
while(m--)
{
scanf("%d%d%d%d",&type,&x,&y,&c);
if(type == )printf("%d\n",query(,x,y,c));
else update(,x,y,type,c);
}
}
return ;
}
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <queue>
#include <math.h>
#define LL long long
using namespace std;
const LL MAX = 1e6 + ;
LL INF = 1e8;
LL MOD = ; LL PowerMod(LL a, LL b)
{
LL ans = ;
a = a % MOD;
while(b > ) {
if(b % == )
ans = (ans * a) % MOD;
b = b / ;
a = (a * a) % MOD;
}
return ans;
} LL a[MAX];
LL lazy[MAX << ][]; void PushDown(LL rt){
if(lazy[rt][] != -){
lazy[rt << ][] = lazy[rt << | ][] = lazy[rt][] % MOD;
lazy[rt << ][] = lazy[rt << | ][] = ;
lazy[rt << ][] = lazy[rt << | ][] = ;
lazy[rt][] = -;
} if(lazy[rt][] != ){
if(lazy[rt << ][] != -){
lazy[rt << ][] *= lazy[rt][];
lazy[rt << ][] %= MOD;
} else{
PushDown(rt << );
lazy[rt << ][] *= lazy[rt][];
lazy[rt << ][] %= MOD;
} if(lazy[rt << | ][] != -){
lazy[rt << | ][] *= lazy[rt][];
lazy[rt << | ][] %= MOD;
} else{
PushDown(rt << | );
lazy[rt << | ][] *= lazy[rt][];
lazy[rt << | ][] %= MOD;;
}
lazy[rt][] = ;
} if(lazy[rt][] != ){
if(lazy[rt << ][] != -){
lazy[rt << ][] += lazy[rt][];
lazy[rt << ][] %= MOD;
} else{
PushDown(rt << );
lazy[rt << ][] += lazy[rt][];
lazy[rt << ][] %= MOD;
} if(lazy[rt << | ][] != -){
lazy[rt << | ][] += lazy[rt][];
lazy[rt << | ][] %= MOD;
} else{
PushDown(rt << | );
lazy[rt << | ][] += lazy[rt][];
lazy[rt << | ][] %= MOD;
}
lazy[rt][] = ;
}
} void Build(LL l, LL r, LL rt){
lazy[rt][] = -;
lazy[rt][] = ;
lazy[rt][] = ;
if(l == r){
lazy[rt][] = ;
return ;
}
LL m = (l + r) >> ;
Build(l, m, rt << );
Build(m + , r, rt << | );
} LL L, R, C;
void Update0(LL l, LL r, LL rt){
if(L <= l && r <= R){
lazy[rt][] = C;
lazy[rt][] %= MOD;
lazy[rt][] = ;
lazy[rt][] = ;
return ;
}
PushDown(rt);
LL m = (l + r) >> ;
if(L <= m){
Update0(l, m, rt << );
}
if(R > m){
Update0(m + , r, rt << | );
}
} void Update1(LL l, LL r, LL rt){
if(L <= l && r <= R){
if(lazy[rt][] != -){
lazy[rt][] *= C;
lazy[rt][] %= MOD;
} else{
PushDown(rt);
lazy[rt][] *= C;
lazy[rt][] %= MOD;
}
return ;
} LL m = (l + r) >> ;
PushDown(rt);
if(L <= m){
Update1(l, m, rt << );
}
if(R > m){
Update1(m + , r, rt << | );
}
} void Update2(LL l, LL r, LL rt){
if(L <= l && r <= R){
if(lazy[rt][] != -){
lazy[rt][] += C;
lazy[rt][] %= MOD;
} else{
PushDown(rt);
lazy[rt][] += C;
lazy[rt][] %= MOD;
}
return ;
} LL m = (l + r) >> ;
PushDown(rt);
if(L <= m){
Update2(l, m, rt << );
}
if(R > m){
Update2(m + , r, rt << | );
}
} LL ans = ;
void Query(LL l, LL r, LL rt){
if(L <= l && r <= R && lazy[rt][] != -){
ans = ans + (r - l + ) * PowerMod(lazy[rt][], C);
ans %= MOD;
return ;
} PushDown(rt);
LL m = (l + r) >> ;
if(L <= m){
Query(l, m, rt << );
}
if(R > m){
Query(m + , r, rt << | );
}
}
int main(int argc, char const *argv[])
{
LL n, m; while(){
scanf("%lld%lld", &n, &m);
if(n == && m == ){
break;
}
Build(, n, );
while(m--){
LL op;
scanf("%lld%lld%lld%lld", &op, &L, &R, &C);
if(op == ){
Update2(, n, );
} else if(op == ){
Update1(, n, );
} else if(op == ){
Update0(, n, );
} else{
ans = ;
Query(, n, );
printf("%lld\n", ans);
}
}
}
return ;
}
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