试题链接

题意：

求出一个１到ｎ的排列，这个排列只告诉了你从第二个数字开始的比这个数字小的数的个数。

题解：

如果一个数p，p的前面有m个比他小的数，后面有n个比他小的数，那p的位置就是n+m+1
我们可以二分这个p，如果m+n+1>mid,mid就取小了；否则就取大了
p前面有多少比他小的数，在输入答案时就知道了
p后面有多少比他小的数，可以用树状数组来搞定
我们在处理时应该倒序处理，想想最后一个数的真实值就是a[n]+1,然后删除最后一个数，看倒数第二个，一次类推即可

代码：

# include<cstdio>
# include<iostream>
# include<fstream>
# include<algorithm>
# include<functional>
# include<cstring>
# include<string>
# include<cstdlib>
# include<iomanip>
# include<numeric>
# include<cctype>
# include<cmath>
# include<ctime>
# include<queue>
# include<stack>
# include<list>
# include<set>
# include<map>using namespace std;const double PI=4.0*atan(1.0);typedef long long LL;
typedef unsigned long long ULL;# define inf 999999999
# define MAX 8000+4int a[MAX];
int ans[MAX];
int tree[MAX];int n;
int lowbit(int x)
{return x&(-x);
}
int getsum ( int pos )//后面比他小的数字的个数 
{int ans = 0;while ( pos > 0 ){ans+=tree[pos];pos-=lowbit(pos);}return ans;
}void update ( int pos,int val )
{while ( pos <= n ){tree[pos]+=val;pos+=lowbit(pos);}
}int my_search ( int k )
{int left = 1, right = n;while ( left < right ){int mid = ( left+right )>>1;int num = getsum (mid);if ( mid - 1 < num+k ){left = mid+1;}else{right = mid;}}return left;
}int main(void)
{while ( scanf("%d",&n)!=EOF ){memset(tree,0,sizeof(tree));memset(ans,0,sizeof(ans));a[1] = 0;for ( int i = 2;i <= n;i++ )scanf("%d",&a[i]);for ( int i = n;i >= 1;i-- ){int k = my_search(a[i]);update(k,1);//每确定一个位置，进行更新 ans[i] = k;}for ( int i = 1;i <= n;i++ )printf("%d\n",ans[i]);}return 0;
}