Find all combinations that sum to n

Given a number n

Requirements:
(1) The value of the integer on the left side of the equation is 1~n-1.
(2) It is required that the sum of the left side of the equation is n.

<code>若 n = 3;
1 + 1 + 1 = 3;
1 + 2 = 3;</code>

Reply content:

Given a number n

Requirements:
(1) The value of the integer on the left side of the equation is 1~n-1.
(2) It is required that the sum of the left side of the equation is n.

<code>若 n = 3;
1 + 1 + 1 = 3;
1 + 2 = 3;</code>

The poster can go and learn the general function

This should be a template question for the parent function

<code>function calcN (n) {
    var res = [],
        cache = {};

    loop(n);

    function loop(k, arr) {
        arr = arr || [];
        var i = 1, count = k / 2 | 0;
        cache[k] = true;
        while (i <= count) {
            
            res.push(arr.concat([i, k - i]));

            if (!cache.hasOwnProperty(i)) {
                loop(i, [k - i].concat(arr));
            }

            if (!cache.hasOwnProperty(k - i)) {
                loop(k - i, [i].concat(arr));
            }

            i++;        
        }
  }
  return res;
}
// 测试部分：
console.log(calcN(5));
// 输出
[ [ 1, 4 ],
  [ 1, 1, 3 ],
  [ 1, 1, 1, 2 ],
  [ 1, 1, 1, 1, 1 ],
  [ 1, 2, 2 ],
  [ 2, 3 ] ]</code>

For C++, just modify it based on the above, use map and vector