Little Girl and Maximum Sum CodeForces - 276C

Posted 2020-09-30 啦啦啦

---恢复内容开始---

The little girl loves the problems on array queries very much.

One day she came across a rather well-known problem: you‘ve got an array of n elements (the elements of the array are indexed starting from 1); also, there are q queries, each one is defined by a pair of integers l_i, r_i (1?≤?l_i?≤?r_i?≤?n). You need to find for each query the sum of elements of the array with indexes from l_i to r_i, inclusive.

The little girl found the problem rather boring. She decided to reorder the array elements before replying to the queries in a way that makes the sum of query replies maximum possible. Your task is to find the value of this maximum sum.

Input

The first line contains two space-separated integers n (1?≤?n?≤?2·10⁵) and q (1?≤?q?≤?2·10⁵) — the number of elements in the array and the number of queries, correspondingly.

The next line contains n space-separated integers a_i (1?≤?a_i?≤?2·10⁵) — the array elements.

Each of the following q lines contains two space-separated integers l_i and r_i (1?≤?l_i?≤?r_i?≤?n) — the i-th query.

Output

In a single line print a single integer — the maximum sum of query replies after the array elements are reordered.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Example

Input

3 3
5 3 2
1 2
2 3
1 3

Output

25

Input

5 3
5 2 4 1 3
1 5
2 3
2 3

Output

33

题意：给一个数列，和q次询问，在回答询问之前，对数列重新排一次序，使得q次询问的和最大。
题解：先计算出所有询问中每个下标出现的次数，出现次数多的就把数列中的较大值分配给它。重点在于计算下标出现的次数！

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<string>
 4 #include<cstring>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 typedef long long ll;
 9 
10 const int maxn=200005;
11 
12 int n,m;
13 int a[maxn],q[maxn],s[maxn];
14 
15 int main()
16 {   int l,r;
17     cin>>n>>m;
18     for(int i=1;i<=n;i++){
19         q[i]=0,s[i]=0;
20         scanf("%d",&a[i]);
21     }
22     for(int i=1;i<=m;i++){          //
23         scanf("%d%d",&l,&r);
24         q[l]++,q[r+1]--;
25     }                                  一个叫差分数列的东西
26     s[0]=0;
27     for(int i=1;i<=n;i++)
28         s[i]=s[i-1]+q[i];           //
29     sort(a+1,a+n+1);
30     sort(s+1,s+n+1);
31     ll sum=0;
32     for(int i=1;i<=n;i++) sum+=(long long)a[i]*s[i];    //不用long long 的话可能溢出！
33     cout<<sum<<endl;
34 }

 

---恢复内容结束---