Programming thinking week5 assignment B-TT's Magic Cat

topic

One day, the magic cat decided to investigate TT’s ability by giving a problem to him. That is select n cities from the world map, and a[i] represents the asset value owned by the i-th city.
Then the magic cat will perform several operations. Each turn is to choose the city in the interval [l,r] and increase their asset value by c. And finally, it is required to give the asset value of each city after q operations.

Input

The first line contains two integers n,q(1≤n,q≤2⋅10⁵) — the number of cities and operations.
The second line contains elements of the sequence a: integer numbers a1,a2,…,an (−10⁶≤ai≤10⁶).
Then q lines follow, each line represents an operation. The i-th line contains three integers l,r and c (1≤l≤r≤n,−10⁵≤c≤10⁵) for the i-th operation.

Output

Print n integers a1,a2,…,an one per line, and ai should be equal to the final asset value of the i-th city.

Sample Input

4 2
-3 6 8 4
4 4 -2
3 3 1

Sample Output

-3 6 9 2

Ideas

The meaning of the question seems simple, but if you directly modify the points in the interval one by one, the complexity is O (qn), and the timeout.
Since all points in a range are operated, the prefix sum and difference can be used to convert the original array into a difference array and change the interval modification into point modification.
The new array b [i] = a [i] -a [i-1] (i> = 1), then b [] records the difference between two adjacent items in a []. Also a [1] may also be modified, so it willa [0] is assigned as 0As a basis for restoring b [] to a [], it remains unchanged.
Then when modifying the value of the point between the interval [l, r], just modifyb[l],b[r+1], which isDifference between a [l] and a [l-1]as well asDifference between a [r + 1] and a [r].
When b [] is turning a [], a [i] = a [i-1] + b [i].

Code

#include <cstdio>
#include <algorithm>
using namespace std;
long long a[200005],b[200005];//范围q*n最大为2*10^10

int main() {
    int n,q;
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)//从1开始，留出a[0]=0，a[1]要与a[0]查分，要有一个固定的开始值
        scanf("%lld",&a[i]);
    for(int i=1;i<=n;i++)
        b[i]=a[i]-a[i-1];
    for(int i=0;i<q;i++){
        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++)
        a[i]=a[i-1]+b[i];
    for(int i=1;i<=n;i++)
        printf("%lld ",a[i]);
}

to sum up

Pay attention to the data range. Long long should be used. In the future, you need to roughly calculate the data range obtained and use the appropriate type.
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