ZOJ 3745 Salary Increasing

Description

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 nintegers 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}.

这绝对是个水题，但同时又是个不折不扣的大坑题，一开始没注意r_i < l_{i+1 ，差点用线段树做。}

 #include <iostream>
 #include <cstring>
 using namespace std;

 long long num[],n,m,l,r,add;//坑！开100005不够，因为工资一开始最高为100000，但后面可以涨工资。
 int main()
 {

     while (cin>>n>>m)
     {
           memset(num,,sizeof(num));
           for (int i=;i<=n;i++)
           {
               cin>>add;
               num[add]++;
           }
           for (int i=;i<=m;i++)
           {
               cin>>l>>r>>add;
               for (int j=r;j>=l;j--)//坑！应该从大到小计算，不然会对后面产生影响。
               {
                   if (num[j])
                   {
                              num[j+add]+=num[j];
                              num[j]=;
                   }
               }
           }
           add=;
           for (int i=;i<;i++)//坑！一开始只从1算到100000，wrong了好几次，后来发现工资可能超过100000.
               add=add+num[i]*i;
           cout <<add<<endl;
     }
     return ;
 }

#include <iostream>#include <cstring>using namespace std;
long long num[600100],n,m,l,r,add;//坑！开100005不够，因为工资一开始最高为100000，但后面可以涨工资。 int main(){        while (cin>>n>>m)    {          memset(num,0,sizeof(num));          for (int i=1;i<=n;i++)          {              cin>>add;              num[add]++;          }          for (int i=1;i<=m;i++)          {              cin>>l>>r>>add;              for (int j=r;j>=l;j--)//坑！应该从大到小计算，不然会对后面产生影响。               {                  if (num[j])                  {                             num[j+add]+=num[j];                             num[j]=0;                  }              }          }          add=0;          for (int i=1;i<600100;i++)//坑！一开始只从1算到100000，wrong了好几次，后来发现工资可能超过100000.               add=add+num[i]*i;          cout <<add<<endl;    }    return 0;}