bzoj5055: 膜*（BIT）

　　大水题WA了两发T T

　　记录一下a[i]的前缀和，a[i]*a[j]就是sigma(a[j]*sumi[j-1])

　　记录一下a[i]*a[j]的前缀和，a[i]*a[j]*a[k]就是sigma(a[k]*sumij[k-1])

　　因为要求ai<aj<ak，所以前缀和必须用权值树状数组来统计

#include<iostream> 
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath> 
#include<algorithm> 
#define MOD(x) ((x)>=mod?(x)-mod:(x))
using namespace std;
const int maxn=500010,mod=19260817;
int n,N,ans;
int a[maxn],b[maxn],lisan[maxn],tree[2][maxn];
char buf[80000010],*ptr=buf-1;
inline int read()
{
    char c=*++ptr;int s=0,t=1;
    while(c<48||c>57)c=*++ptr;
    while(c>=48&&c<=57){s=s*10+c-'0';c=*++ptr;}
    return s*t;
}
inline int lowbit(int x){return x&-x;}
inline void add(int x,int delta,int ty){for(;x<=N;x+=lowbit(x))tree[ty][x]=MOD(tree[ty][x]+delta);}
inline int query(int x,int ty){int sum=0;for(;x;x-=lowbit(x))sum=MOD(sum+tree[ty][x]);return sum;}
int main()
{
    fread(buf,1,sizeof(buf),stdin);n=read();
    for(int i=1;i<=n;i++)a[i]=read(),a[i]%=mod,b[i]=a[i];N=n;
    sort(b+1,b+1+N);N=unique(b+1,b+1+N)-b-1;
    for(int i=1;i<=n;i++)lisan[i]=lower_bound(b+1,b+1+N,a[i])-b;
    for(int i=1;i<=n;i++)
    {
        ans=(ans+1ll*a[i]*query(lisan[i]-1,1))%mod;
        int x=1ll*a[i]*query(lisan[i]-1,0)%mod;
        add(lisan[i],a[i],0);add(lisan[i],x,1);
    }
    printf("%d
",ans);
}