sdnoj1072（模拟）

1072.我们爱递归
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 ……;

}

}

去年10月份做的这道题，错了很多遍，就放到了现在，刚刚又把这道题翻出来，在去年写的代码的输出上加了一个空格，竟然就ac了？？？？？！！！！！
好搞笑啊哈哈哈
纯递归！开心！

#include<cstdio>
#include<iostream>
using namespace std;
int plus1(int a)
{
   
    printf("[plus1(%d) output]Begin invoke plus1(%d)\n",a,a);
    if(a==1)
    {
   
        printf("[plus1(1) output]return plus1(1) = 1\n");
        return 1;
    }
    else
    {
   
        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);
        int c=plus1(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,a+c);
        return a+c;
    }
}
int main()
{
   
    int n;
    scanf("%d",&n);
    printf("%d\n",plus1(n));
    return 0;
}