N (1 <= N <= 100,000) monkeys in the mountains, conveniently numbered 1..N, are once again standing in a row. Monkey i has height Hi(1 <= Hi<= 1,000,000).
Each monkey is looking to his left toward those with higher index numbers. We say that monkey i "looks up" to monkey j if i < j and Hi< Hj. For each monkey i, we would like to know the index of the first monkey in line looked up to by monkey i.
Input
Input consists of several testcases. The format of each case as follow:
Line 1: A single integer: NLines 2..N+1: Line i+1 contains the single integer: Hi
Output
For each testcase, output N lines. Line i contains a single integer representing the smallest index of a monkey up to which monkey i looks. If no such monkey exists, print 0.
Sample Input
6326112
Sample Output
330660
Hint
Monkey 1 and 2 both look up to monkey 3; monkey 4 and 5 both look up to monkey 6; and monkey 3 and 6 do not look up to any monkey.
题意：告诉n个一排猴子，每个猴子向右看，问第一个比自己高的猴子的位置，，没有就输出0，最后一个也是0
#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <queue>#include <cmath>#include <algorithm>#include <stack>#define N 100009using namespace std;int a[N];int main(){int n;while(~scanf("%d",&n)){for(int i=1;i<=n;i++){scanf("%d",&a[i]);}int ans[N];memset(ans,0,sizeof ans);ans[n]=n;//初始化为n，但是实际值为0for(int i=n-1;i>=1;i–)//找右边第一个比a[i]小的位置{int tt=i;while(tt<n&&a[i]>=a[tt+1]) tt=ans[tt+1];ans[i]=tt;}for(int i=1;i<n;i++){if(ans[i]==n)//当是最后一个猴子的时候，并不一定保证比当前猴子高if(a[i]<a[n])printf("%d\n",ans[i]+1);elsecout<<0<<endl;elseprintf("%d\n",ans[i]+1);}cout<<0<<endl;}return 0;}
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累死累活不说，走马观花反而少了真实体验，
