HDU 4578 Transformation （线段树）
题意：
4种操作：
区间加和， 区间赋值，区间乘法， 查询区间 一次方和，二次方和，三次方和。
思路：
区间赋值优先级最高。
直接把加法标记和乘法标记直接赋0， 然后计算和。
其次是乘法， 先计算和， 如果发现有加法标记，在更新一下加法标记， 因为 (a+b)*x = a*x + b*x; 相当于原数乘以x 在加bx ，此时加法标记由b变成了bx。
最后是加法。
更新区间和时， 算一下递推关系即可。
智障错误， wa了两遍= =
#include
#include
#include
using namespace std;
const int maxn = 100000 + 10;
const int mod = 10007;
int n, q;
struct node{
int l,r;
int sum[3];
int addr;
int setv;
int mult;
}nod[maxn<<2];
void build(int l,int r,int o){
nod[o].l = l;
nod[o].r = r;
for (int i = 0; i < 3; ++i){
nod[o].sum[i] = 0;
}
nod[o].addr = nod[o].setv = 0;
nod[o].mult = 1;
if (l == r) return;
int m = l + r >> 1;
build(l,m,o<<1);
build(m+1,r,o<<1|1);
}
void pushup(int o){
int lson = o << 1;
int rson = o << 1 | 1;
for (int i = 0; i < 3; ++i){
nod[o].sum[i] = (nod[lson].sum[i] + nod[rson].sum[i]) % mod;
// printf("%d %d %d\n", nod[lson].sum[i], nod[rson].sum[i], nod[o].sum[i]);
}
}
void uset(int o,int c,int l,int r){
nod[o].setv = c % mod;
nod[o].addr = 0;
nod[o].mult = 1;
nod[o].sum[2] = (((((r-l+1)%mod)*c)%mod*c)%mod*c)%mod;
nod[o].sum[1] = ((((r-l+1)%mod)*c)%mod*c)%mod;
nod[o].sum[0] = ((r-l+1)%mod*c)%mod;
}
void uadd(int o,int c,int l,int r){
nod[o].addr = (nod[o].addr + c) % mod;
nod[o].sum[2] = (nod[o].sum[2] + ( ( ((3*c*c)%mod *nod[o].sum[0])%mod + (3*c*nod[o].sum[1])%mod)%mod + (((r-l+1)%mod *c)%mod *c)%mod *c) % mod) % mod;
nod[o].sum[1] = (nod[o].sum[1] + ( (((((r-l+1)%mod)*c)%mod *c)%mod + (2*c*nod[o].sum[0])%mod ) )%mod) % mod;
nod[o].sum[0] = (nod[o].sum[0] + ((r-l+1)%mod * c))%mod;
}
void umul(int o,int c,int l,int r){
nod[o].mult = (nod[o].mult * c) %mod;
nod[o].sum[2] = (nod[o].sum[2] * ( ((c*c)%mod *c)%mod )) % mod ;
nod[o].sum[1] = (nod[o].sum[1] * ((c*c)%mod) )%mod;
nod[o].sum[0] = (nod[o].sum[0] * c) % mod;
if (nod[o].addr){
nod[o].addr = (nod[o].addr * c) % mod;
}
}
void pushdown(int o){
int l = nod[o].l;
int r = nod[o].r;
int m = l + r >> 1;
int lson = o << 1;
int rson = o << 1 | 1;
if (nod[o].setv != 0){
uset(lson, nod[o].setv, l, m);
uset(rson, nod[o].setv, m+1, r);
nod[o].setv = 0;
}
if (nod[o].mult != 1){
umul(lson, nod[o].mult, l, m);
umul(rson, nod[o].mult, m+1, r);
nod[o].mult = 1;
}
if (nod[o].addr != 0){
uadd(lson, nod[o].addr, l, m);
uadd(rson, nod[o].addr, m+1, r);
nod[o].addr = 0;
}
}
void update(int L,int R,int c,int l,int r,int o,int way){
if (L <= l && r <= R){
if (way == 3){
uset(o,c,l,r);
// printf("%d %d %d o = %d\n",l,r, nod[o].sum[1],o);
return;
}
else if (way == 2){
umul(o,c,l,r);
}
else {
uadd(o,c,l,r);
}
return ;
}
pushdown(o);
int m = l + r >> 1;
if (m >= L){
update(L,R,c,l,m,o<<1,way);
}
if (m < R){
update(L,R,c,m+1,r,o<<1|1,way);
}
pushup(o);
// printf("pushup: %d %d %d o = %d\n",l,r, nod[o].sum[1],o);
}
int query(int L,int R,int c,int l,int r,int o){
if (L <= l && r <= R){
return nod[o].sum[c];
}
int m = l + r >> 1;
int ans = 0;
pushdown(o);
if (m >= L){
ans = (ans + query(L,R,c, l, m, o << 1)) % mod;
}
if (m < R){
ans = (ans + query(L,R,c,m+1, r, o<<1|1)) % mod;
}
pushup(o);
return ans;
}
int main(){
while(~scanf("%d %d",&n, &q) && (n || q)){
int op,x,y,c;
build(1,n,1);
while(q--){
scanf("%d %d %d %d",&op, &x, &y, &c);
if (op != 4){
update(x,y,c,1,n,1,op);
}
else {
// printf("%d\n",nod[1].sum[1]);
printf("%d\n", query(x,y,c-1,1,n,1));
}
}
}
return 0;
}
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子科技大学！
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TransformationTime Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)Total Submission(s): 5450 Accepted Submission(s): 1325
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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