SDNU 1072.我们爱递归（水题）

Description

我们想明白递归函数到底是怎么运作的，所以想用递归来计算当给定一个正整数n(n < 16)时，求出n + (n-1)……+1的值,同时进行一些输出
要求写一个int plus1(int a)形式的函数，在进入递归函数时，输出相关信息。在调用结束返回值前，输出相关信息。最后输出n + (n-1)……+1的值

Input

3

Output

[plus1(3) output]Begin invoke plus1(3)
[plus1(3) output]I want to calculate plus1(3), require 3 and the value of plus1(2),so I need to invoke plus1(2)
[plus1(2) output]Begin invoke plus1(2)
[plus1(2) output]I want to calculate plus1(2), require 2 and the value of plus1(1),so I need to invoke plus1(1)
[plus1(1) output]Begin invoke plus1(1)
[plus1(1) output]return plus1(1) = 1
[plus1(2) output]I got the value of plus1(1), and plus it to 2then return3
[plus1(3) output]I got the value of plus1(2), and plus it to 3then return6
6

Sample Input

3

Sample Output

[plus1(3) output]Begin invoke plus1(3)
[plus1(3) output]I want to calculate plus1(3), require 3 and the value of plus1(2),so I need to invoke plus1(2)
[plus1(2) output]Begin invoke plus1(2)
[plus1(2) output]I want to calculate plus1(2), require 2 and the value of plus1(1),so I need to invoke plus1(1)
[plus1(1) output]Begin invoke plus1(1)
[plus1(1) output]return plus1(1) = 1
[plus1(2) output]I got the value of plus1(1), and plus it to 2then return3
[plus1(3) output]I got the value of plus1(2), and plus it to 3then return6
6

Hint

int plus1(int a)
{
    // 从这里输出开始调用时的信息
    if(a == 1)
    {
        // 从这里输出最后一次调用的返回信息
        return ……;
    }
    else
    {
        // 从这里输出调用时的信息
        // 开始进行递归调用下一步
        ……
        int c = plus1(a - 1);
        // 从这里输出返回值前的信息
        ……
        return ……;
    }
}

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long

int n, sum;

int plu(int a)
{
    printf("[plus1(%d) output]Begin invoke plus1(%d)\n", a, a);
    if(a != 1)
    {
        printf("[plus1(%d) output]I want to calculate plus1(%d), require %d and the value of plus1(%d),so I need to invoke plus1(%d)\n", a, a, a, a-1, a-1);
        sum = a+plu(a-1);
        printf("[plus1(%d) output]I got the value of plus1(%d), and plus it to %dthen return%d\n", a, a-1, a, sum);
        return sum;
    }
    else
    {
        printf("[plus1(1) output]return plus1(1) = 1\n");
        return 1;
    }
}

int main()
{
    scanf("%d", &n);
    plu(n);
    printf("%d\n", sum);
}