BZOJ 4597 随机序列

一定要想到，对于一个空位如果填了+，那么一定有一个表达式这里填-号使得后面的全部抵消掉。这点十分重要。

于是发现这个答案只和前缀积有关，线段树维护即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100500
#define mod 1000000007
using namespace std;
long long n,q,table[maxn],a[maxn],x,y;
long long root,tot=0,ls[maxn<<2],rs[maxn<<2],sum1[maxn<<2],sum2[maxn<<2];
void get_table()
{
    table[0]=1;
    for (long long i=1;i<=100000;i++)
        table[i]=(table[i-1]*3)%mod;
}
void pushup(long long now,long long left,long long right)
{
    long long mid=(left+right)>>1;
    long long ret1,ret2,ret3;
    ret1=((sum1[ls[now]]-sum2[ls[now]])+mod);
    ret2=sum2[ls[now]];
    ret3=sum1[rs[now]];
    sum1[now]=(((ret1*table[right-mid])%mod+((ret2*table[right-mid-1])%mod*2)%mod)%mod+(ret3*ret2)%mod)%mod;
    sum2[now]=(sum2[ls[now]]*sum2[rs[now]])%mod;
}
void build(long long &now,long long left,long long right)
{
    now=++tot;
    if (left==right)
    {
        sum1[now]=sum2[now]=a[left]%mod;
        return;
    }
    long long mid=(left+right)>>1;
    build(ls[now],left,mid);
    build(rs[now],mid+1,right);
    pushup(now,left,right);
}
void modify(long long now,long long left,long long right,long long pos,long long x)
{
    if (left==right)
    {
        sum1[now]=sum2[now]=x%mod;
        return;
    }
    long long mid=(left+right)>>1;
    if (pos<=mid) modify(ls[now],left,mid,pos,x);
    else modify(rs[now],mid+1,right,pos,x);
    pushup(now,left,right);
}
int main()
{
    scanf("%lld%lld",&n,&q);
    for (long long i=1;i<=n;i++)
        scanf("%lld",&a[i]);
    get_table();
    build(root,1,n);
    for (long long i=1;i<=q;i++)
    {
        scanf("%lld%lld",&x,&y);
        modify(root,1,n,x,y);
        printf("%lld\n",sum1[root]%mod);
    }
    return 0;
}