BZOJ 1798 Ahoi 2009 维护序列seq
来源:程序员人生 发布时间:2014-09-30 03:41:45 阅读次数:3388次
题目大意:维护一个序列,能够区间加,区间乘,然后取区间和模一个数的值。
思路:线段树维护一个有两个域的标记,一个表示加,一个表示乘。下传的时候一起下传,先乘后加。
CODE:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 100010
#define MO p
#define LEFT (pos << 1)
#define RIGHT (pos << 1|1)
#define CNT (r - l + 1)
using namespace std;
long long p;
struct Mark{
long long p_mark,m_mark;
Mark() {
m_mark = 1;
p_mark = 0;
}
void operator +=(long long c) {
p_mark = (p_mark + c) % MO;
}
void operator *=(long long c) {
p_mark = (p_mark * c) % MO;
m_mark = (m_mark * c) % MO;
}
void operator +=(const Mark &a) {
m_mark = (m_mark * a.m_mark) % MO;
p_mark = (p_mark * a.m_mark + a.p_mark) % MO;
}
};
struct Complex{
long long sum;
Mark mark;
}tree[MAX << 2];
int cnt,asks;
long long src[MAX];
void BuildTree(int l,int r,int pos);
void Mulitiply(int l,int r,int x,int y,int pos,long long c);
void Plus(int l,int r,int x,int y,int pos,long long c);
int Ask(int l,int r,int x,int y,int pos);
inline void PushDown(int pos,int cnt);
int main()
{
cin >> cnt >> p;
for(int i = 1;i <= cnt; ++i)
scanf("%lld",&src[i]);
BuildTree(1,cnt,1);
cin >> asks;
for(int flag,i = 1;i <= asks; ++i) {
scanf("%d",&flag);
int x,y,z;
switch(flag) {
case 1:
scanf("%d%d%d",&x,&y,&z);
Mulitiply(1,cnt,x,y,1,z);
break;
case 2:
scanf("%d%d%d",&x,&y,&z);
Plus(1,cnt,x,y,1,z);
break;
case 3:
scanf("%d%d",&x,&y);
printf("%d
",(int)Ask(1,cnt,x,y,1));
break;
}
}
return 0;
}
void BuildTree(int l,int r,int pos)
{
tree[pos].mark.p_mark = 0;
tree[pos].mark.m_mark = 1;
if(l == r) {
tree[pos].sum = src[l] % MO;
return ;
}
int mid = (l + r) >> 1;
BuildTree(l,mid,LEFT);
BuildTree(mid + 1,r,RIGHT);
tree[pos].sum = (tree[LEFT].sum + tree[RIGHT].sum) % MO;
}
void Mulitiply(int l,int r,int x,int y,int pos,long long c)
{
if(l == x && y == r) {
tree[pos].sum = (tree[pos].sum * c) % MO;
tree[pos].mark *= c;
return ;
}
PushDown(pos,CNT);
int mid = (l + r) >> 1;
if(y <= mid) Mulitiply(l,mid,x,y,LEFT,c);
else if(x > mid) Mulitiply(mid + 1,r,x,y,RIGHT,c);
else {
Mulitiply(l,mid,x,mid,LEFT,c);
Mulitiply(mid + 1,r,mid + 1,y,RIGHT,c);
}
tree[pos].sum = (tree[LEFT].sum + tree[RIGHT].sum) % MO;
}
void Plus(int l,int r,int x,int y,int pos,long long c)
{
if(l == x && y == r) {
tree[pos].sum = (tree[pos].sum + (c * CNT) % MO) % MO;
tree[pos].mark += c;
return ;
}
PushDown(pos,CNT);
int mid = (l + r) >> 1;
if(y <= mid) Plus(l,mid,x,y,LEFT,c);
else if(x > mid) Plus(mid + 1,r,x,y,RIGHT,c);
else {
Plus(l,mid,x,mid,LEFT,c);
Plus(mid + 1,r,mid + 1,y,RIGHT,c);
}
tree[pos].sum = (tree[LEFT].sum + tree[RIGHT].sum) % MO;
}
int Ask(int l,int r,int x,int y,int pos)
{
if(l == x && y == r)
return tree[pos].sum;
PushDown(pos,CNT);
int mid = (l + r) >> 1;
if(y <= mid) return Ask(l,mid,x,y,LEFT);
if(x > mid) return Ask(mid + 1,r,x,y,RIGHT);
int left = Ask(l,mid,x,mid,LEFT);
int right = Ask(mid + 1,r,mid + 1,y,RIGHT);
return (left + right) % MO;
}
inline void PushDown(int pos,int cnt)
{
if(tree[pos].mark.m_mark == 1 && !tree[pos].mark.p_mark)
return ;
tree[LEFT].mark += tree[pos].mark;
tree[RIGHT].mark += tree[pos].mark;
if(tree[pos].mark.m_mark != 1) {
tree[LEFT].sum = (tree[LEFT].sum * tree[pos].mark.m_mark) % MO;
tree[RIGHT].sum = (tree[RIGHT].sum * tree[pos].mark.m_mark) % MO;
tree[pos].mark.m_mark = 1;
}
if(tree[pos].mark.p_mark) {
tree[LEFT].sum = (tree[LEFT].sum + ((cnt - (cnt >> 1)) * tree[pos].mark.p_mark) % MO) % MO;
tree[RIGHT].sum = (tree[RIGHT].sum + ((cnt >> 1) * tree[pos].mark.p_mark) % MO) % MO;
tree[pos].mark.p_mark = 0;
}
}
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