Shoi2016 random sequence

To give you a series, insert the plus sign, minus sign or multiplication sign between two adjacent numbers.

Each time a single point of modification is supported, the sum of all the expressions obtained in this way is obtained. The membrane is 1e9 + 7.

SOL:

I am a Sb...

It can be found that if a plus sign appears at a position, there must be a minus sign to remove it, so the answer is the sum of prefix products that appear several times.

Calculate the contribution of each prefix product.

#include<bits/stdc++.h>#define int long longusing namespace std;inline int read(){    int x = 0,f = 1;char ch = getchar();    for(;!isdigit(ch);ch = getchar())if(ch == ‘-‘)f = -f;    for(;isdigit(ch);ch = getchar())x = 10 * x + ch - ‘0‘;    return x * f;}int n,q;const int mod = 1e9 + 7,maxn = 1e5 + 10;int a[maxn],fac[maxn];#define ls (x << 1)#define rs ((x << 1) | 1)int seg[maxn << 2],tag[maxn << 2];inline int pw(int x,int t){    int res = 1;x %= mod;    while(t)    {        if(t & 1)res = res * x % mod;        x = x * x % mod;        t = t >> 1;    }    return res;}inline void build(int x,int l,int r){    tag[x] = 1;    if(l == r)seg[x] = fac[l];    else    {        int mid = (l + r) >> 1;        build(ls,l,mid);build(rs,mid + 1,r);        seg[x] = (seg[ls] + seg[rs]) % mod;    }    }inline void pushdown(int x,int l,int r){    if(tag[x] != 1)    {        (tag[ls] *= tag[x]) %= mod;(tag[rs] *= tag[x]) %= mod;        (seg[ls] *= tag[x]) %= mod;(seg[rs] *= tag[x]) %= mod;        tag[x] = 1;    }}inline void update(int x,int l,int r,int L,int R,int val){    if(L <= l && r <= R)    {        (seg[x] *= val) %= mod;        (tag[x] *= val) %= mod;        return;    }    pushdown(x,l,r);    int mid = (l + r) >> 1;    if(L <= mid)update(ls,l,mid,L,R,val);    if(R > mid)update(rs,mid + 1,r,L,R,val);    seg[x] = (seg[ls] + seg[rs]) % mod;}signed main(){    n = read(),q = read();    for(int i=1;i<=n;i++)a[i] = read();    int mul = 1;    for(int i=1;i<=n;i++)    {        mul = (long long)mul * a[i] % mod;        if(i == n)fac[i] = mul;        else fac[i] = (long long)mul * 2 * pw(3, n - i - 1) % mod;    }build(1,1,n);    while(q--)    {        int p = read(),v = read();        update(1,1,n,p,n,(long long)v * pw(a[p],mod - 2) % mod);        a[p] = v;        printf("%lld\n",seg[1]);    }}

View code

$ \ Sum _ {I = 1} ^ {n-1} sum_ I \ times 2 \ times 3 ^ {n-i-1} + sum_n $

$ Sum $ array is prefix Product

Shoi2016 random sequence