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[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;
}

 

#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;}