基础算法：一维差分

题目描述

参考代码

/*
给定a1, a2, a3, ..., an
构造差分数组b[N]，使得
ai = b1 + b2 + ... + b[i]
核心操作：将a[L~R]全部加上C，等价于b[L] += C, b[R+1] -= C
*/

#include <iostream>

using namespace std;

const int N = 100010;

int n, m;
int a[N], b[N];

void insert(int l, int r, int c)
{
    b[l] += c;
    b[r + 1] -= c;
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    
    for (int i = 1; i <= n; i++) insert(i, i, a[i]);
    
    while (m--)
    {
        int l, r, c;
        cin >> l >> r >> c;
        insert(l, r, c);
    }
    
    for (int i = 1; i <= n; i++) a[i] = a[i - 1] + b[i];
    
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    
}