ACM-ICPC LA 4329 Ping pong (tree-like array)

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&itemid=8&page=show_problem& problem=2330

Reference: "An introduction to Algorithms Classic Training Guide" Rujia P197

The book above is written on the main topic, the problem-solving ideas are written out.

I put my own in here.

AC Code:

1#include <stdio.h>2#include <string.h>3 4 #definePeople 200015 #defineVALUE 1000016 7typedefLong LongLL;8 9 intN, A[people], amin[people], amax[people], tree[value];Ten  One  A intLowbit (intx) { -     returnX &-x; - } the  - voidAddintXintd) { -      while(X <VALUE) { -TREE[X] + =D; +X + =lowbit (x); -     } + } A  at intSumintx) { -     intRET =0; -      while(X >0){ -RET + =Tree[x]; -X-=lowbit (x); -     } in     returnret; - } to  + intMain () { -     intT; thescanf"%d", &t); *      while(t--){ $scanf"%d", &n);Panax Notoginseng          for(inti =1; I <= N; ++i) { -scanf"%d", &a[i]); the         } +  Amemset (Tree,0,sizeof(tree)); the          for(inti =1; I <= N; ++i) { +Amin[i] = SUM (a[i]-1);//I personal statistics on the left of the first person and the total number of skill values less than -Amax[i] = i-1-Amin[i];//statistics skill value is greater than first person and total number of people on the left $Add (A[i],1); $         } -  -LL ans =0; thememset (Tree,0,sizeof(tree)); -          for(inti = n; I >=1; --i) {WuyiLL bmin = SUM (a[i]-1);//The total number of statistics skill values less than the first person and on the right of the first person theAns + = Bmin * Amax[i] + (N-i-bmin) *Amin[i]; -Add (A[i],1); Wu         } -printf"%lld\n", ans); About     } $     return 0; -}

ACM-ICPC LA 4329 Ping pong (tree-like array)