【xsy1504】 pitcure 树状数组

数据范围：$n≤2\times 10^5$

以下是题解：

 #include<bits/stdc++.h>
 #define L unsigned int
 #define MOD 16777216
 #define lowbit(x) ((x)&(-(x)))
 #define M 200005
 using namespace std;

 L b[M]={},a[M]={},n;
 void add(L x,L k){for(L i=x;i<=n;i+=lowbit(i)) b[i]+=k;}
 L query(L x){L k=; for(L i=x;i;i-=lowbit(i)) k+=b[i]; return k;}

 L solve1(){
     static L r[M]={},j[M]={}; L res=;
     memset(b,,sizeof(b));
     for(L i=n;i;i--){
         r[i]=query(n)-query(a[i]);
         add(a[i],);
     }
     memset(b,,sizeof(b));
     for(L i=;i<=n;i++){
         j[i]=query(a[i]);
         add(a[i],i);
     }
     memset(b,,sizeof(b));
     for(L i=;i<=n;i++){
         L l=query(a[i]),now=;
         now+=1LL*l*(i-);
         now-=1LL*l*(l-)/;
         now-=j[i];
         res+=1LL*now*r[i];
         add(a[i],);
     }
     return res;
 }
 L solve2(){
     static L r[M]={},j[M]={}; L res=;
     memset(b,,sizeof(b));
     for(L i=n;i;i--){
         r[i]=query(n)-query(a[i]);
         add(a[i],);
     }
     memset(b,,sizeof(b));
     for(L i=;i<=n;i++){
         j[i]=query(a[i]);
         add(a[i],a[i]);
     }
     memset(b,,sizeof(b));
     for(L i=;i<=n;i++){
         L l=query(a[i]),now=;
         now+=1LL*l*(a[i]-);
         now-=1LL*l*(l-)/;
         now-=j[i];
         res+=1LL*now*r[i];
         add(a[i],);
     }
     return res;
 }

 L solve3(){
     L res=; memset(b,,sizeof(b));
     for(L i=n;i;i--){
         L hh=query(n)-query(a[i]);
         res+=(1LL*hh*(hh-)*(hh-)/);
         add(a[i],);
     }
     return res;
 }
 L solve4(){
     static L f[M]={};
     for(L i=n;i;i--) f[i]=;
     for(L x=;x<=;x++){
         memset(b,,sizeof(b));
         for(L i=n;i;i--){
             L hh=query(n)-query(a[i]);
             add(a[i],f[i]);
             f[i]=hh;
         }
     }
     L res=; for(L i=;i<=n;i++) res+=f[i];
     return res;
 }

 main(){
     scanf("%u",&n);
     for(L i=;i<=n;i++) scanf("%u",a+i);
     L res1=solve1();//1x2x
     L res2=solve2();//14xx
     L res3=solve3();//1xxx
     L res4=solve4();//
     printf("%u\n",((res1+res2-res3+res4)%MOD+MOD)%MOD);
 }