P2023 [AHOI2009]维护序列 题解（线段树）

题目链接

P2023 [AHOI2009]维护序列

解题思路

线段树板子。不难，但是...有坑。坑有多深？一页(WA)。
由于乘法可能乘(k=0)，我这种做法可能会使结果产生负数。于是就有了这篇题解。
（详情见代码注释）

AC代码

#include<stdio.h>
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
typedef long long ll;
int n,m;
ll mod,k,a[500010];
struct Tree{
	int left,right;
	ll data,lazy,mul;
}tree[2000010];
void build(int p,int left,int right){
	tree[p].left=left;
	tree[p].right=right;
	tree[p].mul=1;
	if(left==right){tree[p].data=a[left];return;}
	build(p<<1,left,(left+right)>>1);
	build(p<<1|1,((left+right)>>1)+1,right);
	tree[p].data=(tree[p<<1].data+tree[p<<1|1].data)%mod;
}
void pushdown(int p){
	ll mul=tree[p].mul,lazy=tree[p].lazy;
	tree[p<<1].lazy*=mul;
	tree[p<<1].lazy+=lazy;tree[p<<1].lazy%=mod;
	tree[p<<1].mul*=mul;tree[p<<1].mul%=mod;
	tree[p<<1|1].lazy*=mul;
	tree[p<<1|1].lazy+=lazy;tree[p<<1|1].lazy%=mod;
	tree[p<<1|1].mul*=mul;tree[p<<1|1].mul%=mod;
	tree[p].data*=tree[p].mul;
	tree[p].data+=(tree[p].right-tree[p].left+1)*tree[p].lazy;
	tree[p].data%=mod;
	tree[p].lazy=0;tree[p].mul=1;
}
void add(int left,int right,ll k,int p){
	int l=tree[p].left,r=tree[p].right;
	if(l>right||r<left||p>4*n)return;
	pushdown(p);
	if(l>=left&&r<=right){
		tree[p].lazy+=k;
		tree[p].lazy%=mod;
		return;
	}
	tree[p].data+=k*(min(right,r)-max(left,l)+1);
	tree[p].data%=mod;
	add(left,right,k,p<<1);
	add(left,right,k,p<<1|1);
}
ll multy(int left,int right,ll k,int p){
	int l=tree[p].left,r=tree[p].right;
	if(l>right||r<left||p>4*n)return 0;
	pushdown(p);
	if(l>=left&&r<=right){
		ll temp=tree[p].data*tree[p].mul+tree[p].lazy*(r-l+1);
		tree[p].lazy*=k;tree[p].lazy%=mod;
		tree[p].mul*=k;tree[p].mul%=mod;
		return ((k-1)*temp%mod+mod)%mod;//非常重要！！！！！！ 
	}
	ll temp=multy(left,right,k,p<<1)+multy(left,right,k,p<<1|1); 
	tree[p].data+=temp;
	tree[p].data=;
	return temp;
}
ll query(int left,int right,int p){
	int l=tree[p].left,r=tree[p].right;
	if(l>right||r<left||p>4*n)return 0;
	pushdown(p);
	if(l>=left&&r<=right)return tree[p].data;
	return query(left,right,p<<1)+query(left,right,p<<1|1);
}
int main(){
	int s,x,y,i;
	scanf("%d%lld",&n,&mod);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
	build(1,1,n);
	scanf("%d",&m);
	while(m--){
		scanf("%d%d%d",&s,&x,&y);
		if(s-3){
			scanf("%lld",&k);
			if(s-1)add(x,y,k,1);
			else multy(x,y,k,1);
		}else printf("%lld
",query(x,y,1)%mod);
	}
	return 0;
}