程序设计思维与实践 Week5 作业 B TT's Magic Cat

题目描述：

Thanks to everyone's help last week, TT finally got a cute cat. But what TT didn't expect is that this is a magic cat.

One day, the magic cat decided to investigate TT's ability by giving a problem to him. That is select nn cities from the world map, and a[i]a[i] represents the asset value owned by the ii-th city.

Then the magic cat will perform several operations. Each turn is to choose the city in the interval [l,r][l,r] and increase their asset value by cc. And finally, it is required to give the asset value of each city after qq operations.

Could you help TT find the answer?

简述：给出n给数，没有q次操作。每次操作给出一个区间[l,r]，对于该区间中的每个元素，加上c，输出最终n个数的值

input：

The first line contains two integers n,qn,q (1≤n,q≤2⋅105)(1≤n,q≤2·105) — the number of cities and operations.

The second line contains elements of the sequence aa: integer numbers a1,a2,...,ana1,a2,...,an (−106≤ai≤106)(−106≤ai≤106).

Then qq lines follow, each line represents an operation. The ii-th line contains three integers l,rl,r and cc (1≤l≤r≤n,−105≤c≤105)(1≤l≤r≤n,−105≤c≤105) for the ii-th operation.

简述：第一行包含两个整数n，q。第2行到第q+1行，每行三个整数：l r c，表示区间[l,r]，以及要加上的数c

output：

Print nn integers a1,a2,…,ana1,a2,…,an one per line, and aiai should be equal to the final asset value of the ii-th city.

简述：输出最终每个数的值

思路：

经典的差分题，对于n个数a[1]~a[n]。令b[1]=a[1]，b[i]=a[i]-a[i-1]。则b的前缀和为对应的a[i]

#include<iostream>
using namespace std;
long long ans,a[300010],b[300010];
int n,q,l,r,c;
int main()
{
  cin>>n>>q;
  for(int i=1;i<=n;i++)
  {
    cin>>a[i];
    if(i==1)
      b[i]=a[i];
    else
      b[i]=a[i]-a[i-1];
  }
  for(int i=1;i<=q;i++)
  {
    cin>>l>>r>>c;
    b[l]+=c;
    b[r+1]-=c;
  }
  for(int i=1;i<=n;i++)
  {
    ans+=b[i];
    cout<<ans<<" ";
  }
  return 0;
}