luogu 4927 [1007]梦美与线段树 概率与期望 + 线段树

考场上切了不考虑没有逆元的情况(出题人真良心). 

考场代码： 

#include <cstdio> 
#include <algorithm>  
#define lson (now<<1) 
#define rson (now<<1|1) 
#define ll long long
#define setIO(s) freopen(s".in","r",stdin) , freopen(s".out","w",stdout)     
using namespace std;   
char *p1, *p2, buf[100000];
namespace IO 
{
    #define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ ) 
    int rd() {
    int x = 0, f = 1;
    char c = nc();
    while (c < 48) {
        if (c == '-')
            f = -1;
        c = nc();
    }
    while (c > 47) {
        x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
    }
    return x * f;
    }
}; 
const int mod=998244353,N=120005; 
int arr[N],n,Q; 
inline ll qpow(ll base,ll k) {
    ll tmp=1;
    for(;k;base=base*base%mod,k>>=1)if(k&1)tmp=tmp*base%mod;
    return tmp;
} 
inline ll inv(ll k) {
    return qpow(k,mod-2);
}                        
struct Node {
    int len;
    ll sum,sqr,sumlen,sqrlen,lazy;        
}t[N<<2];  
inline void pushup(int l,int r,int now) {
    int mid=(l+r)>>1; 
    t[now].sum=t[lson].sum; 
    t[now].sqr=t[lson].sqr; 
    t[now].sumlen=t[lson].sumlen;   
    t[now].sqrlen=t[lson].sqrlen; 
    if(r>mid)  {
        t[now].sum=(t[now].sum+t[rson].sum)%mod; 
        t[now].sqr=(t[now].sqr+t[rson].sqr)%mod; 
        t[now].sumlen=(t[now].sumlen+t[rson].sumlen)%mod;    
        t[now].sqrlen=(t[now].sqrlen+t[rson].sqrlen)%mod; 
    }
    t[now].sqr=(t[now].sqr+(ll)t[now].sum*t[now].sum)%mod;   
    t[now].sumlen=(t[now].sumlen+t[now].len*t[now].sum%mod)%mod; 
    t[now].sqrlen=(t[now].sqrlen+(ll)t[now].len*t[now].len%mod)%mod; 
} 
inline void mark(int l,int r,int now,ll v) 
{
    t[now].lazy+=v, t[now].lazy%=mod;   
    t[now].sqr=(t[now].sqr+((v*v)%mod)*t[now].sqrlen%mod+2ll*v*t[now].sumlen%mod)%mod;   
    t[now].sumlen=(t[now].sumlen+(v*t[now].sqrlen)%mod)%mod;     
    t[now].sum=(t[now].sum+(t[now].len*v)%mod)%mod;        
} 
inline void pushdown(int l,int r,int now) 
{
    int mid=(l+r)>>1; 
    if(t[now].lazy) 
    {
        mark(l,mid,lson,t[now].lazy); 
        if(r>mid) mark(mid+1,r,rson,t[now].lazy); 
        t[now].lazy=0; 
    }
}
void build(int l,int r,int now) {
    t[now].len=r-l+1;  
    if(l==r) {
        t[now].sum=arr[l]; 
        t[now].sqr=(ll)arr[l]*arr[l]%mod;                               
        t[now].sumlen=t[now].len*t[now].sum%mod;  
        t[now].sqrlen=(ll)t[now].len*t[now].len%mod; 
        return; 
    }
    int mid=(l+r)>>1;  
    if(l<=mid) build(l,mid,lson); 
    if(r>mid) build(mid+1,r,rson);   
    pushup(l,r,now); 
} 
void update(int l,int r,int now,int L,int R,ll v) {
    if(l>=L&&r<=R) {
        mark(l,r,now,v); 
        return; 
    }
    pushdown(l,r,now); 
    int mid=(l+r)>>1; 
    if(L<=mid) update(l,mid,lson,L,R,v); 
    if(R>mid) update(mid+1,r,rson,L,R,v);   
    pushup(l,r,now);    
}
int main() {
    using namespace IO;   
    int i,j,cas; 
    // setIO("b"); 
    n=rd(),Q=rd(); 
    for(i=1;i<=n;++i) arr[i]=rd(); 
    build(1,n,1);    
    for(cas=1;cas<=Q;++cas) { 
        int opt,l,r,v; 
        opt=rd(); 
        if(opt==1) {
            l=rd(),r=rd(),v=rd(), update(1,n,1,l,r,v); 
        }
        if(opt==2) {
            ll a=t[1].sqr,b=t[1].sum; 
            printf("%lld\n",a*inv(b)%mod);  
        }
    }
    return 0; 
}

标签：4927,int,luogu,线段,sumlen,sqr,now,sum,mod
来源： https://www.cnblogs.com/guangheli/p/11393873.html