OJ Problem Set - 3745

Salary Increasing

Edward has established a company with n staffs. He is such a kind man that he did Q times salary increasing for his staffs. Each salary increasing was described by three integers (l, r, c). That means Edward will add c units money for the staff whose salaxy is in range [l, r] now. Edward wants to know the amount of total money he should pay to staffs after Q times salary increasing.

Input

The input file contains multiple test cases.

Each case begin with two integers : n -- which indicate the amount of staff; Q -- which indicate Q times salary increasing. The following n integers each describes the initial salary of a staff(mark as a_i). After that, there are Q triples of integers (l_i, r_i, c_i) (i=1..Q) which describe the salary increasing in chronological.

1 ≤ n ≤ 10⁵ , 1 ≤ Q ≤ 10⁵ , 1 ≤ a_i ≤ 10⁵ , 1 ≤ l_i ≤ r_i ≤ 10⁵ , 1 ≤ c_i ≤ 10⁵ , r_i < l_i+1

Process to the End Of File.

Output

Output the total salary in a line for each case.

Sample Input

4 1

1 2 3 4

2 3 4

Sample Output

18

Hint

{1, 2, 3, 4} → {1, 4, 6, 7}.

Author: CHEN, Weijie

Contest: ZOJ Monthly, December 2013

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<algorithm>

#include<map>

#include<queue>

#include<set>

#include<stack>

#include<cmath>

#include<vector>

#define inf 0x3f3f3f3f

#define Inf 0x3FFFFFFFFFFFFFFFLL

#define eps 1e-9

#define pi acos(-1.0)

using namespace std;

typedef long long ll;

const int maxn=+;

int num[maxn];

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int n,q;

    while(~scanf("%d%d",&n,&q))

    {

        int tmp;

        ll sum=;

        memset(num,,sizeof(num));

        for(int i=;i<n;++i)

        {

            scanf("%d",&tmp);

            num[tmp]++;

            sum+=tmp;

        }

        int l,r,c;

        while(q--)

        {

            scanf("%d%d%d",&l,&r,&c);

            //if(l>r) swap(l,r);

            for(int i=r;i>=l;--i)

            {

                sum+=(ll)c*num[i];

                if(i+c<maxn) num[i+c]+=num[i];

                num[i]=;

            }

        }

        printf("%lld\n",sum);

    }

    return ;

}