题目传送门:https://codeforces.com/problemset/problem/276/C
题目大意:
给定一串长度为(n)的序列(A),(m)个询问((l_i,r_i)),记(V_i=sumlimits_{j=l_i}^{r_i}A_j)
问,在对(A)序列任意排序后,所能得到(sumlimits_{i=1}^mV_i)最大是多少
统计每个点被询问的次数,将大数尽可能填入询问次数多的位置即可
至于统计次数,就不需要数据结构了,差分即可
/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
int f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
template<typename T>inline T read(T x){
int f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=2e5;
int A[N+10],S[N+10];
int main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n=read(0),m=read(0);
for (int i=1;i<=n;i++) A[i]=read(0);
for (int i=1;i<=m;i++){
int l=read(0),r=read(0);
S[l]++,S[r+1]--;
}
for (int i=1;i<=n;i++) S[i]+=S[i-1];
sort(A+1,A+1+n);
sort(S+1,S+1+n);
ll Ans=0;
for (int i=1;i<=n;i++) Ans+=1ll*A[i]*S[i];
printf("%lld
",Ans);
return 0;
}