1 solutions

  • 0
    @ 2025-12-1 17:43:40
    #include<bits/stdc++.h>
    using namespace std;
    const int N=100010;
    int a[N],b[N]; //b[i]表示原来数组的差分数组
    /*
    差分数组 
    b[1]=a[1];
    b[2]=a[2]-a[1];
    .....
    b[i]=a[i]-a[i-1]
    
    对原始数组进行a[l]...a[r]这个区域+c,等价于b[l]+=c,b[r+1]-=c 
    
    前缀和数组
    a[i]=b[1]+b[2]....b[i] 
    */ 
    
    
    int main()
    {
    	int n,m;
    	scanf("%d%d",&n,&m);
    	for(int i=1;i<=n;i++) //计算差分数组 
    	{
    		scanf("%d",&a[i]);
    		b[i]=a[i]-a[i-1];
    	}
    	while(m--) //区域进行增减 
    	{
    		int l,r,c;
    		scanf("%d%d%d",&l,&r,&c);
    		b[l]+=c;
    		b[r+1]-=c;
    	}
    	for(int i=1;i<=n;i++) //计算进行 
    	{
    		a[i]=a[i-1]+b[i];
    	}
    	for(int i=1;i<=n;i++)
    	{
    		printf("%d ",a[i]);
    	}
    	return 0;
    }
    
    
    • 1

    Information

    ID
    197
    Time
    500ms
    Memory
    256MiB
    Difficulty
    3
    Tags
    # Submissions
    43
    Accepted
    17
    Uploaded By