AtCoder Beginner Contest 357-F
Problem
同步于博客
Problem
You are given sequences of length \(N\), \(A=(A_1,A_2,\ldots,A_N)\) and \(B=(B_1,B_2,\ldots,B_N)\).
You are also given \(Q\) queries to process in order.
There are three types of queries:
1 l r x: Add \(x\) to each of \(A_l, A_{l+1}, \ldots, A_r\).2 l r x: Add \(x\) to each of \(B_l, B_{l+1}, \ldots, B_r\).3 l r: Print the remainder of \(\displaystyle\sum_{i=l}^r (A_i\times B_i)\) when divided by \(998244353\).
给你两个长度为 \(N\) 的序列 \(A=(A_1,A_2,\ldots,A_N)\) 和 \(B=(B_1,B_2,\ldots,B_N)\) 的序列。
需要按顺序处理 \(Q\) 个查询。
查询有三种类型：
1 l r x: 在 \(A_l, A_{l+1}, \ldots, A_r\) 中的每一条添加 \(x\) 。2 l r x: 向 \(B_l, B_{l+1}, \ldots, B_r\) 中的每一条添加 \(x\) 。3 l r: 打印 \(\displaystyle\sum_{i=l}^r (A_i\times B_i)\) 除以 \(998244353\) 的余数。
Constraints
- \(1\leq N,Q\leq 2\times 10^5\)
- \(0\leq A_i,B_i\leq 10^9\)
- \(1\leq l\leq r\leq N\)
- \(1\leq x\leq 10^9\)
- 全是整数
Solution
看到 \(N,Q\) 的取值范围，发现需要在 \(n\log n\) 的复杂度内解决。值域 \(1\le x\le 10^9\) 也不好做文章。
数列\(A,B\)在\([l,r]\)上在进行操作1 l r x和2 l r y后会变为
\[\begin{align}
&\sum_{i=l}^r(A_i+x)\times (B_i+y)\\
&=\sum_{i=l}^rA_i\times B_i+x\sum_{i=l}^r B_i+y\sum_{i=l}^r A_i+xy(r-l+1)
\end{align}
\]
故我们使用线段树维护三个量 \(\sum A_i\times B_i,\sum A_i,\sum B_i\) 和lazy标记 \(Add_a,Add_b\)
当执行1 l r x 时
- \(\sum A_i\times B_i\) 增加 \(x\sum B_i+x\cdot Add_b\cdot(r-l+1)\)
- \(\sum A_i\) 增加 \(x(r-l+1)\)
- \(Add_a\) 增加 \(x\)
当执行2 l r y 时
- \(\sum A_i\times B_i\) 增加 \(y\sum A_i+y\cdot Add_a\cdot(r-l+1)\)
- \(\sum B_i\) 增加 \(y(r-l+1)\)
- \(Add_b\) 增加 \(y\)
Code
#define P 998244353ll
#define N 800010
struct Edge{
LL l,r,sumA,sumB,sum,addA,addB;
#define l(x) tree[x].l
#define r(x) tree[x].r
#define sumA(x) tree[x].sumA
#define sumB(x) tree[x].sumB
#define sum(x) tree[x].sum
#define addA(x) tree[x].addA
#define addB(x) tree[x].addB
}tree[N];
LL n,m;
LL A[N],B[N];
void build(LL l,LL r,LL p)
{
l(p)=l,r(p)=r;
if(l==r){sumA(p)=A[l]%P;sumB(p)=B[l]%P;sum(p)=sumA(p)*sumB(p)%P;return;}
LL mid=(l+r)>>1;
build(l,mid,p<<1);
build(mid+1,r,p<<1|1);
sum(p)=sum(p<<1)+sum(p<<1|1);
sumA(p)=sumA(p<<1)+sumA(p<<1|1);
sumB(p)=sumB(p<<1)+sumB(p<<1|1);
sum(p)%=P;
sumA(p)%=P;
sumB(p)%=P;
}
void spread(LL p)
{
if(addA(p)||addB(p))
{
addA(p<<1)+=addA(p);
addA(p<<1|1)+=addA(p);
addB(p<<1)+=addB(p);
addB(p<<1|1)+=addB(p);
sum(p<<1)+=addA(p)*sumB(p<<1)+addB(p)*sumA(p<<1)+addA(p)*addB(p)%P*(r(p<<1)-l(p<<1)+1);
sum(p<<1|1)+=addA(p)*sumB(p<<1|1)+addB(p)*sumA(p<<1|1)+addA(p)*addB(p)%P*(r(p<<1|1)-l(p<<1|1)+1);
sumA(p<<1)+=addA(p)*(r(p<<1)-l(p<<1)+1);
sumB(p<<1)+=addB(p)*(r(p<<1)-l(p<<1)+1);
sumA(p<<1|1)+=addA(p)*(r(p<<1|1)-l(p<<1|1)+1);
sumB(p<<1|1)+=addB(p)*(r(p<<1|1)-l(p<<1|1)+1);
addA(p)=addB(p)=0;
sum(p<<1)%=P;
sum(p<<1|1)%=P;
sumA(p<<1)%=P;
sumA(p<<1|1)%=P;
sumB(p<<1)%=P;
sumB(p<<1|1)%=P;
addA(p<<1)%=P;
addB(p<<1)%=P;
addA(p<<1|1)%=P;
addB(p<<1|1)%=P;
}
}
void change(LL l,LL r,LL p,LL k,bool isA)
{
if(l<=l(p)&&r>=r(p)){
if(isA) addA(p)+=k,sum(p)+=k*sumB(p),sumA(p)+=k*(r(p)-l(p)+1);
else addB(p)+=k,sum(p)+=k*sumA(p),sumB(p)+=k*(r(p)-l(p)+1);
sumA(p)%=P;sumB(p)%=P;sum(p)%=P;
addA(p)%=P;addB(p)%=P;
return;
}
LL mid=l(p)+r(p)>>1;
spread(p);
if(l<=mid) change(l,r,p<<1,k,isA);
if(r>mid) change(l,r,p<<1|1,k,isA);
sum(p)=sum(p<<1)+sum(p<<1|1);
sumA(p)=sumA(p<<1)+sumA(p<<1|1);
sumB(p)=sumB(p<<1)+sumB(p<<1|1);
sum(p)%=P;
sumA(p)%=P;
sumB(p)%=P;
}
LL ask(LL l,LL r,LL p)
{
if(l<=l(p)&&r>=r(p)) return sum(p);
spread(p);
LL mid=l(p)+r(p)>>1,val=0;
if(l<=mid) val+=ask(l,r,p<<1);
val%=P;
if(r>mid) val+=ask(l,r,p<<1|1);
val%=P;
return val;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(10);
int q;
cin>>n>>q;
for(int i=1;i<=n;i++) cin>>A[i];
for(int i=1;i<=n;i++) cin>>B[i];
build(1,n,1);
while(q--)
{
int op,l,r,x;
cin>>op>>l>>r;
if(op!=3) cin>>x,change(l,r,1,x,(op==1));
else cout<<ask(l,r,1)<<endl;
}
return 0;
}
Attention
开LL，多取模。
