B. Queue
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
codeforces.com/problemset/problem/91/B
Description
The i-th walrus becomes displeased if there's a younger walrus standing in front of him, that is, if exists such j (i < j), that ai > aj. The displeasure of the i-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the i-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.
The airport manager asked you to count for each of n walruses in the queue his displeasure.
Input
The first line contains an integer n (2 ≤ n ≤ 105) — the number of
walruses in the queue. The second line contains integers ai
(1 ≤ ai ≤ 109).
Note that some walruses can have the same age but
for the displeasure to emerge the walrus that is closer to the head of
the queue needs to be strictly younger than the other one.
Output
Print n numbers: if the i-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the i-th walrus's displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.
Sample Input
6
10 8 5 3 50 45
Sample Output
2 1 0 -1 0 -1
HINT
题意
给你一个数列，让你找到最右边比这个数小的数的位置，如果没有就输出-1
题解：
对与某一位i,假设最右边比他大的数为j，那么a[i]>a[j]，然后将min{a[j+1],a[j+2],...,a[n]}记作min[j+1], 则a[i]<=min[j+1],
不难想到不等式： min[j]<=a[j]<a[i]<=min[j+1]
这就说明min数组单增，然后在单增序列里面找一个值。
这不就是二分吗？然后就没了。
代码
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define N 100050 5 int n,a[N],mi[N]; 6 template<typename T>void read(T&x) 7 { 8 ll k=0; char c=getchar(); 9 x=0; 10 while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar(); 11 if (c==EOF)exit(0); 12 while(isdigit(c))x=x*10+c-'0',c=getchar(); 13 x=k?-x:x; 14 } 15 void read_char(char &c) 16 {while(!isalpha(c=getchar())&&c!=EOF);} 17 int ef(int x,int l,int r) 18 { 19 if (l==r)return l; 20 int mid=(l+r+1)>>1; 21 if (x>mi[mid]) 22 return ef(x,mid,r); 23 else return ef(x,l,mid-1); 24 } 25 int main() 26 { 27 #ifndef ONLINE_JUDGE 28 freopen("aa.in","r",stdin); 29 #endif 30 read(n); 31 for(int i=1;i<=n;i++)read(a[i]); 32 mi[n]=a[n]; 33 for(int i=n-1;i>=1;i--)mi[i]=min(mi[i+1],a[i]); 34 for(int i=1;i<=n;i++) 35 { 36 int ans=ef(a[i],i,n)-i-1; 37 printf("%d ",ans); 38 } 39 }