【P1471 方差】 区间加 区间求平均数 区间求方差
P1471
区间加 区间求平均数 区间求方差
其实这道题的关键在于化方差式子 你把每一项都列开就能得到最后的方差式子为
(a[l]*a[l]+a[l+1]*a[l+1]+…a[r]*a[r])/(r-l+1) - ((a[l]+a[l+1]+…+a[r])/(r-l+1))^2
然后我们只要维护一个 a[i]的平方 维护一个 a[i]
然后用线段树去操作就可以实现了
/*
if you can't see the repay
Why not just work step by step
rubbish is relaxed
to ljq
*/
#include #define dbg3(x1,x2,x3) cout<<#x1<<" = "<#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)
typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod = (int)1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 100025;
double s[MAX_N<<2],s_mul[MAX_N<<2],col[MAX_N<<2];
void up(int rt)
{
s[rt] = s[rt<<1] + s[rt<<1|1];
s_mul[rt] = s_mul[rt<<1] + s_mul[rt<<1|1];
}
void down(int rt,int l,int r)
{
if(col[rt])
{
col[rt<<1] += col[rt];
col[rt<<1|1] += col[rt];
s_mul[rt<<1] += (s[rt<<1])*col[rt]*2+col[rt]*col[rt]*(mid-l+1);
s_mul[rt<<1|1] += (s[rt<<1|1])*col[rt]*2+col[rt]*col[rt]*(r-mid);
s[rt<<1] += (mid-l+1)*col[rt];
s[rt<<1|1] += (r-mid)*col[rt];
col[rt] = 0;
}
}
void build(int rt,int l,int r)
{
s[rt] = s_mul[rt] = col[rt] = 0;
if(l==r)
{
scanf("%lf",&s[rt]);
s_mul[rt] = s[rt]*s[rt];
return ;
}
build(rt<<1,l,mid);
build(rt<<1|1,mid+1,r);
up(rt);
}
void update(int rt,int l,int r,int x,int y,double v)
{
if(x<=l&&r<=y)
{
s_mul[rt]+=v*(s[rt])*2+v*v*(r-l+1);
s[rt]+=(r-l+1)*v;
col[rt]+=v;
return ;
}
down(rt,l,r);
if(x<=mid) update(rt<<1,l,mid,x,y,v);
if(mid<y) update(rt<<1|1,mid+1,r,x,y,v);
up(rt);
}
double query(int rt,int l,int r,int x,int y,int v)
{
if(x<=l&&r<=y)
{
return (v==0?s[rt]:s_mul[rt]);
}
down(rt,l,r);
if(y<=mid) return query(rt<<1,l,mid,x,y,v);
if(mid<x) return query(rt<<1|1,mid+1,r,x,y,v);
return query(rt<<1,l,mid,x,y,v)+query(rt<<1|1,mid+1,r,x,y,v);
}
int main()
{
//ios::sync_with_stdio(false);
//freopen("a.txt","r",stdin);
//freopen("b.txt","w",stdout);
int n,Q,opt,x,y;
double v;
scanf("%d%d",&n,&Q);
build(1,1,n);
while(Q--)
{
scanf("%d%d%d",&opt,&x,&y);
if(opt==1)
{
scanf("%lf",&v);
update(1,1,n,x,y,v);
}
else if(opt==2)
{
double ans = query(1,1,n,x,y,0)/(y-x+1);
printf("%.4f\n",ans);
}
else
{
double ans = query(1,1,n,x,y,1)/(y-x+1) - query(1,1,n,x,y,0)/(y-x+1)*query(1,1,n,x,y,0)/(y-x+1);
printf("%.4f\n",ans);
}
}
//fclose(stdin);
//fclose(stdout);
//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
return 0;
}