BZOJ 1798 (线段树||分块)的标记合并

我原来准备做方差的。。

结果发现不会维护两个标记。。

就是操作变成一个 a*x+b ，每次维护a ， b 即可

加的时候a=1 ，b=v

乘的时候a=v ，b=0

 #include <cstdio>
 const long long Maxn=;

 long long a[Maxn],n,P,l,r,c,m,type;
 struct Node
 {
     long long mul,add,sum,len;
 }tree[Maxn<<];

 inline void Change(long long o,long long mul,long long add)
 {
     tree[o].mul=(tree[o].mul*mul)%P;
     tree[o].add=(tree[o].add*mul+add)%P;
     tree[o].sum=(tree[o].sum*mul+tree[o].len*add)%P;
 }
 inline void push_down(long long o)
 {
     Change(o<<,tree[o].mul,tree[o].add);
     Change(o<<|,tree[o].mul,tree[o].add);
     tree[o].mul=; tree[o].add=;
 }
 inline void push_up(long long o)
 {
     tree[o].sum=(tree[o<<].sum+tree[o<<|].sum)%P;
     tree[o].len=tree[o<<].len+tree[o<<|].len;
 }
 void Build(long long o,long long l,long long r)
 {
     tree[o].mul=; tree[o].add=;
     if (l==r) {tree[o].sum=a[l]; tree[o].len=; return;}
     long long mid=(l+r)>>;
     Build(o<<,l,mid),Build(o<<|,mid+,r);
     push_up(o);
 }
 void Mul(long long o,long long l,long long r,long long p,long long q,long long v)
 {
     if (l==p && r==q)
     {
         Change(o,v,);
         return;
     }
     push_down(o);
     long long mid=(l+r)>>;
     if (q<=mid) Mul(o<<,l,mid,p,q,v);
     if (p>=mid+) Mul(o<<|,mid+,r,p,q,v);
     if (p<=mid && q>=mid+)
         Mul(o<<,l,mid,p,mid,v),Mul(o<<|,mid+,r,mid+,q,v);
     push_up(o);
 }
 void Add(long long o,long long l,long long r,long long p,long long q,long long v)
 {
     if (l==p && r==q)
     {
         Change(o,,v);
         return;
     }
     push_down(o);
     long long mid=(l+r)>>;
     if (q<=mid) Add(o<<,l,mid,p,q,v);
     if (p>=mid+) Add(o<<|,mid+,r,p,q,v);
     if (p<=mid && q>=mid+)
         Add(o<<,l,mid,p,mid,v),Add(o<<|,mid+,r,mid+,q,v);
     push_up(o);
 }
 long long Sum(long long o,long long l,long long r,long long p,long long q)
 {
     if (l==p && r==q) return tree[o].sum;
     long long mid=(l+r)>>;
     push_down(o);
     if (q<=mid) return Sum(o<<,l,mid,p,q);
     if (p>=mid+) return Sum(o<<|,mid+,r,p,q);
     if (p<=mid && q>=mid+)
         return (Sum(o<<,l,mid,p,mid)+Sum(o<<|,mid+,r,mid+,q))%P;
 }
 int main()
 {
     // freopen("c.in","r",stdin);
     scanf("%lld%lld",&n,&P);
     for (long long i=;i<=n;i++) scanf("%lld",&a[i]);
     Build(,,n);
     scanf("%lld",&m);
     for (long long i=;i<=m;i++)
     {
         scanf("%lld",&type);
         if (type==)
         {
             scanf("%lld%lld%lld",&l,&r,&c);
             Mul(,,n,l,r,c);
         }
         if (type==)
         {
             scanf("%lld%lld%lld",&l,&r,&c);
             Add(,,n,l,r,c);
         }
         if (type==)
         {
             scanf("%lld%lld",&l,&r);
             printf("%lld\n",Sum(,,n,l,r));
         }
     }
     return ;
 }

线段树

UPD:2016.6.14

突然发现好久没有写过分块了，想到这道可以分块，push_up写成push_down了..

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <iostream>
 #define LL long long
 using namespace std;
 const LL Maxn=;
 const LL Inf=0x3f3f3f3f;
 LL a[Maxn],n,P,l,r,c,m,type,Block[Maxn];
 struct Node
 {
     LL mul,add,len,sum;
 }tree[Maxn];

 inline void Change(LL o,LL mul,LL add)
 {
     tree[o].mul=(tree[o].mul*mul)%P;
     tree[o].add=(tree[o].add*mul+add)%P;
     tree[o].sum=(tree[o].sum*mul+tree[o].len*add)%P;
 }
 inline void push_down(LL o)
 {
     for (;Block[o]==Block[o-];o--);
     for (LL i=o;Block[o]==Block[i];i++) a[i]=(a[i]*tree[Block[i]].mul+tree[Block[i]].add)%P;
     tree[Block[o]].mul=,tree[Block[o]].add=;
 }

 inline void push_up(LL o)
 {
     tree[Block[o]].sum=;
     for (;Block[o]==Block[o-];o--);
     for (LL i=o;Block[o]==Block[i];i++) tree[Block[o]].sum=(tree[Block[o]].sum+a[i])%P;
 }
 void Mul(LL p,LL q,LL v)
 {
     if (Block[p]==Block[q])
     {
         push_down(p);
         for (LL i=p;i<=q;i++) a[i]=(a[i]*v)%P;
         push_up(p);
         return;
     }
     for (LL i=Block[p]+;i<Block[q];i++) Change(i,v,);
     push_down(p),push_down(q);
     for (LL i=p;Block[i]==Block[p];i++) a[i]=(a[i]*v)%P;
     for (LL i=q;Block[i]==Block[q];i--) a[i]=(a[i]*v)%P;
     push_up(p),push_up(q);
 }

 void Add(LL p,LL q,LL v)
 {
     if (Block[p]==Block[q])
     {
         push_down(p);
         for (LL i=p;i<=q;i++) a[i]=(a[i]+v)%P;
         push_up(p);
         return;
     }
     for (LL i=Block[p]+;i<Block[q];i++) Change(i,,v);
     push_down(p),push_down(q);
     for (LL i=p;Block[i]==Block[p];i++) a[i]=(a[i]+v)%P;
     for (LL i=q;Block[i]==Block[q];i--) a[i]=(a[i]+v)%P;
     push_up(p),push_up(q);
 }
 LL Sum(LL p,LL q)
 {
     LL ret=;
     if (Block[p]==Block[q])
     {
         push_down(p);
         for (LL i=p;i<=q;i++) ret=(ret+a[i])%P;
         return ret;
     }
     for (LL i=Block[p]+;i<Block[q];i++) ret=(ret+tree[i].sum)%P;
     push_down(p),push_down(q);
     for (LL i=p;Block[i]==Block[p];i++) ret=(ret+a[i])%P;
     for (LL i=q;Block[i]==Block[q];i--) ret=(ret+a[i])%P;
     return ret;
 }
 int main()
 {
     scanf("%lld%lld",&n,&P);
     for (LL i=;i<=n;i++) scanf("%lld",&a[i]),a[i]%=P;
     LL pos=(LL)sqrt(n);
     for (LL i=;i<=n;i++) Block[i]=(i-)/pos+;
     memset(tree,,sizeof(tree));
     for (LL i=;i<=n;i++) tree[Block[i]].len++;
     for (LL i=;i<=n;i++) tree[Block[i]].mul=;
     for (LL i=;i<=n;i++) tree[Block[i]].sum=(tree[Block[i]].sum+a[i])%P;

     scanf("%lld",&m);
     for (LL i=;i<=m;i++)
     {
         scanf("%lld",&type);
         if (type==)
         {
             scanf("%lld%lld%lld",&l,&r,&c);
             Mul(l,r,c);
         }
         if (type==)
         {
             scanf("%lld%lld%lld",&l,&r,&c);
             Add(l,r,c);
         }
         if (type==)
         {
             scanf("%lld%lld",&l,&r);
             printf("%lld\n",Sum(l,r));
         }
     }
     return ;
 }

分块