hdu 4578 Transformation 线段树
Transformation
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)Total Submission(s): 3579 Accepted Submission(s): 859
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.
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.
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
线段树：有set, mul,add三个标记。
当set时，把mul，add初始化。
mul操作时，可以把mul*mul, add*mul可以直接合并。然后每个区间先做乘法，然后做加法
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 300000
#define ll long long
int sum[maxn][3];
int mul[maxn],add[maxn],val[maxn];
int lc[maxn],rc[maxn],set[maxn];
int mod = 10007;
int cnt;
void init(){
cnt = 1;
memset(sum,0,sizeof(sum));
memset(lc,0,sizeof(lc));
memset(rc,0,sizeof(rc));
memset(val,0,sizeof(val));
for(int i = 0;i < maxn; i++)
mul[i] = 1;
memset(add,0,sizeof(add));
memset(set,0,sizeof(set));
}
void update(int u,int l,int r){
if(l == r || u == 0) return ;
int len = (r-l+1)%mod;
if(set[u] != 0){
int num = set[u]*set[u]%mod;
sum[u][0] = set[u]*len%mod;
sum[u][1] = num*len%mod;
sum[u][2] = num*set[u]%mod*len%mod;
return ;
}
int s1 = (sum[lc[u]][0] + sum[rc[u]][0])%mod;
int s2 = (sum[lc[u]][1] + sum[rc[u]][1])%mod;
int s3 = (sum[lc[u]][2] + sum[rc[u]][2])%mod;
if(mul[u] > 1){
s1 = (ll)s1*mul[u]%mod;
s2 = (ll)s2*mul[u]*mul[u]%mod;
s3 = (ll)s3*mul[u]*mul[u]*mul[u]%mod;
}
if(add[u] != 0){
ll a1 = add[u],a2= a1*a1,a3=a2*a1;
sum[u][0] = (ll)(s1 + a1*len)%mod;
sum[u][1] = (ll)(s2+2*s1*a1+a2*len)%mod;
sum[u][2] = ((ll)s3+3*s2*a1+3*s1*a2+a3*len)%mod;
}
else sum[u][0] = s1,sum[u][1] = s2,sum[u][2] = s3;
}
void pushdown(int u,int l,int r,int op,int num){
if(u == 0) return ;
if(l == r){
if(op == 1) val[u] = (val[u] + num) % mod;
else if(op == 2) val[u] = val[u] * num % mod;
else if(op == 3) val[u] = num;
sum[u][0] = val[u];
sum[u][1] = (ll)val[u]*val[u]%mod;
sum[u][2] = (ll)val[u]*val[u]*val[u]%mod;
return ;
}
if(op == 3){
set[u] = num;
add[u] = 0;
mul[u] = 1;
update(u,l,r);
return ;
}
int mid = (l+r)/2;
if(set[u] != 0){
pushdown(lc[u],l,mid,3,set[u]);
pushdown(rc[u],mid+1,r,3,set[u]);
set[u] = 0;
}
if(op == 2){
mul[u] = mul[u]*num%mod;
add[u] = add[u]*num%mod;
}
if(op == 1){
if(num == 0){
if(mul[u] != 1){
pushdown(lc[u],l,mid,2,mul[u]);
pushdown(rc[u],mid+1,r,2,mul[u]);
mul[u] = 1;
}
if(add[u] > 0){
pushdown(lc[u],l,mid,1,add[u]);
pushdown(rc[u],mid+1,r,1,add[u]);
}
add[u] = 0;
}
else add[u] = (add[u]+num)%mod;
}
update(u,l,r);
}
void build(int u,int l,int r){
if(l == r) return ;
int mid = (l+r)/2;
lc[u] = cnt++;
rc[u] = cnt++;
build(lc[u],l,mid);
build(rc[u],mid+1,r);
}
int query(int u,int l,int r,int L,int R,int ty){
if(l == L && R == r)
return sum[u][ty];
pushdown(u,l,r,1,0);
int mid = (l+r)/2;
int ans = 0;
if(mid < L) ans = query(rc[u],mid+1,r,L,R,ty);
else if(mid >= R) ans = query(lc[u],l,mid,L,R,ty);
else ans = query(lc[u],l,mid,L,mid,ty)+query(rc[u],mid+1,r,mid+1,R,ty);
update(u,l,r);
return ans%mod;
}
void work(int u,int l,int r,int L,int R,int op,int num){
if(l == L && R == r){
pushdown(u,l,r,op,num);
return ;
}
pushdown(u,l,r,1,0);
int mid = (l+r)/2;
if(mid < L) work(rc[u],mid+1,r,L,R,op,num);
else if(mid >= R) work(lc[u],l,mid,L,R,op,num);
else work(lc[u],l,mid,L,mid,op,num),work(rc[u],mid+1,r,mid+1,R,op,num);
update(u,l,r);
}
int main(){
int n,m;
//freopen("4578.in","r",stdin);
//freopen("4578.out","w",stdout);
while(scanf("%d%d",&n,&m),n+m){
int op,l,r,num;
init();
cnt++;
build(1,1,n);
while(m--){
scanf("%d%d%d%d",&op,&l,&r,&num);
if(op != 4) work(1,1,n,l,r,op,num%mod);
else printf("%d\n",query(1,1,n,l,r,num-1));
}
}
return 0;
}