POJ 2828 segment Tree Jack processing

I'll give you a sequence number. num "i" and the corresponding numerical value output the corresponding value of the final sequence, such as

40 771 511) 332 69

First Step

Part II 77 51

Step three 77 33 51

Part fourth 77 33 69 51

The position that appears after the first is fixed, so it is processed from the back. The number of slots in each node of the segment tree that is stored in the current interval;

#include<stdio.h>#include<string.h>#include<iostream>using namespaceStd;#define LL(X) (X<<1)#define RR(X) ((X<<1)|1)struct node {intCont;}Tree[4*200000];intNum[200010][3],Queue[200010];int deal(intL,intR,intPoint){Tree[Point].Cont=R-L+1;if(L==R)return 0;intMid=(L+R)/2;deal(L,Mid,LL(Point));deal(Mid+1,R,RR(Point));return 0; }int Update(intL,intR,intPos,intPoint,intValue){Tree[Point].Cont--;if(L==R){Queue[L]=Value;return 0;}intMid=(L+R)/2;if(Pos<=Tree[LL(Point)].Cont) Update(L,Mid,Pos,LL(Point),Value);Else Update(Mid+1,R,Pos-Tree[LL(Point)].Cont,RR(Point),Value);return 0;}int Main(){intN,I,J; while(~scanf("%d",&N)){deal(1,N,1); for(I=1;I<=N;I++){scanf("%d%d",&Num[I][1],&Num[I][2]);Num[I][1]++;} for(I=N;I>=1;I--){   Update(1,N,Num[I][1],1,Num[I][2]);} for(I=1;I<=N;I++){if(I!=1) printf(" ");printf("%d",Queue[I]);}printf("\ n");}return 0;}

POJ 2828 segment Tree Jack processing