Frequency

题目描述

Snuke loves constructing integer sequences.

There are N piles of stones, numbered 1 through N. The pile numbered i consists of ai stones.

Snuke will construct an integer sequence s of length Σai, as follows:

Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.

Select a pile with one or more stones remaining, and remove a stone from that pile.

If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.

We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,…,N, how many times does it occur in the lexicographically smallest sequence?



Constraints

1≤N≤105

1≤ai≤109

输入

The input is given from Standard Input in the following format:

N

a1 a2 … aN

输出

Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.

样例输入

2

1 2

样例输出

2

1

分析：偷看题解，慢慢写出来的。。。。

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#define range(i,a,b) for(int i=a;i<=b;++i)
#define LL long long
#define rerange(i,a,b) for(int i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
using namespace std;
int n;
LL ans[100005];
pair<LL,LL>aa[100005];
bool cmp(pair<LL,LL>a,pair<LL,LL>b){
    return a.second>b.second;
}
void init() {
    cin>>n;
    fill(ans,0);
    range(i,1,n){
        aa[i].first=i;
        cin>>aa[i].second;
    }
    sort(aa+1,aa+1+n,cmp);
    aa[n+1].second=0;
}
void solve(){
    LL tmp=1,num=aa[1].first;
    range(i,2,n+1){
        if(aa[i].second==aa[i-1].second)++tmp;
        else ans[num]+=(aa[i-1].second-aa[i].second)*(tmp++);
        if(aa[i].first<num)num=aa[i].first;
    }
    range(i,1,n)printf("%lld\n",ans[i]);
}
int main() {
    init();
    solve();
    return 0;
}

View Code