数据范围：$n≤2 imes 10^5$
以下是题解：
1 #include<bits/stdc++.h> 2 #define L unsigned int 3 #define MOD 16777216 4 #define lowbit(x) ((x)&(-(x))) 5 #define M 200005 6 using namespace std; 7 8 L b[M]={0},a[M]={0},n; 9 void add(L x,L k){for(L i=x;i<=n;i+=lowbit(i)) b[i]+=k;} 10 L query(L x){L k=0; for(L i=x;i;i-=lowbit(i)) k+=b[i]; return k;} 11 12 L solve1(){ 13 static L r[M]={0},j[M]={0}; L res=0; 14 memset(b,0,sizeof(b)); 15 for(L i=n;i;i--){ 16 r[i]=query(n)-query(a[i]); 17 add(a[i],1); 18 } 19 memset(b,0,sizeof(b)); 20 for(L i=1;i<=n;i++){ 21 j[i]=query(a[i]); 22 add(a[i],i); 23 } 24 memset(b,0,sizeof(b)); 25 for(L i=1;i<=n;i++){ 26 L l=query(a[i]),now=0; 27 now+=1LL*l*(i-1); 28 now-=1LL*l*(l-1)/2; 29 now-=j[i]; 30 res+=1LL*now*r[i]; 31 add(a[i],1); 32 } 33 return res; 34 } 35 L solve2(){ 36 static L r[M]={0},j[M]={0}; L res=0; 37 memset(b,0,sizeof(b)); 38 for(L i=n;i;i--){ 39 r[i]=query(n)-query(a[i]); 40 add(a[i],1); 41 } 42 memset(b,0,sizeof(b)); 43 for(L i=1;i<=n;i++){ 44 j[i]=query(a[i]); 45 add(a[i],a[i]); 46 } 47 memset(b,0,sizeof(b)); 48 for(L i=1;i<=n;i++){ 49 L l=query(a[i]),now=0; 50 now+=1LL*l*(a[i]-1); 51 now-=1LL*l*(l-1)/2; 52 now-=j[i]; 53 res+=1LL*now*r[i]; 54 add(a[i],1); 55 } 56 return res; 57 } 58 59 L solve3(){ 60 L res=0; memset(b,0,sizeof(b)); 61 for(L i=n;i;i--){ 62 L hh=query(n)-query(a[i]); 63 res+=(1LL*hh*(hh-1)*(hh-2)/6); 64 add(a[i],1); 65 } 66 return res; 67 } 68 L solve4(){ 69 static L f[M]={0}; 70 for(L i=n;i;i--) f[i]=1; 71 for(L x=1;x<=3;x++){ 72 memset(b,0,sizeof(b)); 73 for(L i=n;i;i--){ 74 L hh=query(n)-query(a[i]); 75 add(a[i],f[i]); 76 f[i]=hh; 77 } 78 } 79 L res=0; for(L i=1;i<=n;i++) res+=f[i]; 80 return res; 81 } 82 83 main(){ 84 scanf("%u",&n); 85 for(L i=1;i<=n;i++) scanf("%u",a+i); 86 L res1=solve1();//1x2x 87 L res2=solve2();//14xx 88 L res3=solve3();//1xxx 89 L res4=solve4();//1234 90 printf("%u ",((res1+res2-res3+res4)%MOD+MOD)%MOD); 91 }