题解 | #【模板】差分#

#include <iostream>
#include <vector>

using namespace std;
using LL = long long;

int main() {
    int n, m;
    cin >> n >> m;

    vector<LL> a(n + 1);
    // 差分数组，大小设为 n+2 以便处理 d[r+1]
    vector<LL> d(n + 2, 0);

    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        // 构建初始差分数组
        d[i] = a[i] - a[i - 1];
    }

    for (int k = 0; k < m; ++k) {
        int l, r;
        LL v;
        cin >> l >> r >> v;
        // 区间修改等价于对差分数组的两个点进行修改
        d[l] += v;
        d[r + 1] -= v;
    }

    // 通过前缀和还原最终数组
    for (int i = 1; i <= n; ++i) {
        a[i] = a[i - 1] + d[i];
    }

    // 输出结果
    for (int i = 1; i <= n; ++i) {
        cout << a[i] << (i == n ? "" : " ");
    }
    cout << endl;

    return 0;
}

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        long[] a = new long[n + 1];
        // 差分数组，大小设为 n+2 以便处理 d[r+1]
        long[] d = new long[n + 2];

        // 构建初始差分数组
        for (int i = 1; i <= n; i++) {
            a[i] = sc.nextLong();
            d[i] = a[i] - a[i - 1];
        }
        
        // 处理m次修改
        for (int k = 0; k < m; k++) {
            int l = sc.nextInt();
            int r = sc.nextInt();
            long v = sc.nextLong();
            // 区间修改等价于对差分数组的两个点进行修改
            d[l] += v;
            d[r + 1] -= v;
        }

        // 通过前缀和还原最终数组并输出
        for (int i = 1; i <= n; i++) {
            a[i] = a[i - 1] + d[i];
            System.out.print(a[i] + (i == n ? "" : " "));
        }
        System.out.println();
    }
}

# 读取 n 和 m
n, m = map(int, input().split())
# 读取原始数组 (0-based)
a = list(map(int, input().split()))

# 差分数组 (1-based for easier logic)
# 大小为 n+2 以便处理 d[r+1]
d = [0] * (n + 2)

# 构建初始差分数组
d[1] = a[0]
for i in range(1, n):
    d[i+1] = a[i] - a[i-1]

# 处理m次修改
for _ in range(m):
    l, r, v = map(int, input().split())
    # 区间修改等价于对差分数组的两个点进行修改
    d[l] += v
    d[r + 1] -= v

# 通过前缀和还原最终数组
result = [0] * n
result[0] = d[1]
for i in range(1, n):
    result[i] = result[i-1] + d[i+1]

# 输出结果
print(*result)