Codeforces 276C Little Girl and Maximum Sum【贪心】

C. Little Girl and Maximum Sum

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 li, ri (1 ≤ li ≤ ri ≤ n). You need to find for each query the sum of elements of the array with indexes from li to ri, 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 ai (1 ≤ ai ≤ 2·105) — the array elements.

Each of the following q lines contains two space-separated integers li and ri (1 ≤ li ≤ ri ≤ 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 %I64dspecifier.

Examples

input

3 35 3 21 22 31 3

output

25

input

5 35 2 4 1 31 52 32 3

output

33

题目大意：

给出N个数，我们可以将其任意排列。

一共有Q个查询，每个查询表示求一个区间的和，问我们如何排列序列使得所有的查询的总和最大。

输出最大总和。

思路：

我们在每个查询区间【L,R】的时候，我们其实就是加了每个位子上的数一次，那么我们考虑计算每个位子的被计算的次数即可。

被计算次数多的，我们将原序列中大的数放置在这个位子上即可。

排排序就行了。。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[1150000];int sum[1150000];int pos[1150000];int main(){    int n,q;    while(~scanf("%d%d",&n,&q))    {        memset(sum,0,sizeof(sum));        for(int i=1;i<=n;i++)scanf("%d",&a[i]);        for(int i=1;i<=q;i++)        {            int x,y;scanf("%d%d",&x,&y);            sum[x]++;            sum[y+1]--;        }        int now=0;        for(int i=1;i<=n;i++)        {            now+=sum[i];            pos[i]=now;        }        sort(pos+1,pos+1+n);        sort(a+1,a+1+n);        __int64 ans=0;        for(int i=n;i>=1;i--)        {            ans+=(__int64)pos[i]*a[i];        }        printf("%I64d\n",ans);    }}