ZOJ3475解题报告

Edward has established a company with n staffs. He is such a kind man that he didQ 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 afterQ 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 followingn integers each describes the initial salary of a staff(mark as a_i). After that, there areQ 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 11 2 3 42 3 4

Sample Output

18

Hint

{1, 2, 3, 4} → {1, 4, 6, 7}. 

/* 题目：Salary Increasing   链接 ： //http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3745   题意：在数组的指定范围的数值的元素上增加的一个值，求最后的数组总值   思路：暴力。先求初始时总值，并且记录每个值的个数，增加值时，在原来的和的基础上增加相应的值，最后就得出总值了*/#include<cstdio>#include<cstring>#include<iostream>using namespace std;#define LL long long#define M 100050LL a[M],l,r,n,c,q,ctotal[M],temp[M],ans;int main(){    LL i,j;    while(~scanf("%lld %lld",&n,&q))    {        ans=0;        memset(ctotal,0,sizeof(ctotal));        memset(temp,0,sizeof(temp));        for(i=1;i<=n;i++)            {                scanf("%lld",&a[i]);                temp[a[i]]++;                ans+=a[i];            }             memcpy(ctotal,temp,sizeof(temp));        while(q--)        {            scanf("%lld %lld %lld",&l,&r,&c);            for(i=l;i<=r;i++)            {                if(i+c<M)                ctotal[i+c]=temp[i]+temp[c+i];                ans+=temp[i]*c;                temp[i]=0;            }            for(i=l;i<l+c&&i<M;i++) ctotal[i]=0;            for(i=l+c;i<=r+c&&i<M;i++) temp[i]=ctotal[i];        }        printf("%lld\n",ans);    }    return 0;}