趁熱來一發線段樹,主要是聯繫一下線段樹的基本操作,寫了兩道題之後確實感覺手感好了很多,這道題的思路是,從後向前挨個數,最後一個牛總是可以判斷是第幾個,然後就把相應的位置佔住,以後再數的時候就不能數已經佔住的位置了,說的不是很清楚,但是代碼的思路比較清晰:
#include <iostream>
#include <cstdio>
#include <memory.h>
using namespace std;
#define maxn 8050
struct node{
int left;
int right;
int len;
};
node tree[maxn<<2]={0};
int cows[maxn]={0};
int ans[maxn]={0};
void build(int left, int right, int loc){
tree[loc].left = left;
tree[loc].right = right;
tree[loc].len = right - left + 1;
if(left == right)
return;
int mid = (right + left)>>1;
build(left,mid,loc<<1);
build(mid+1,right,(loc<<1)+1);
}
int query(int num,int loc){
tree[loc].len -- ;
if(tree[loc].left == tree[loc].right){
return tree[loc].left;
}
if(tree[loc<<1].len < num){
return query(num-tree[loc<<1].len,(loc<<1)+1);
}
else{
return query(num,loc<<1);
}
}
int main(){
int n;
scanf("%d",&n);
for(int i=2;i<=n;++i){
scanf("%d",&cows[i]);
}
build(1,n,1);
for(int i=n;i>0;--i){
ans[i] = query(cows[i]+1,1);
}
for(int i=1;i<=n;++i){
printf("%d\n",ans[i]);
}
//system("pause");
return 0;
}