每天坚持做算法题，熟悉基本的套路，巩固语言基础，总结经验记下所学所想。
题目

HINT

代码

const int maxn = 1000
int ans[maxn], v[maxn];

//满足1/c≤a/b的最小c
int get_first(int a, int b){
    for(int i = 1; ;++i){
        if(1/i <= a/b) return i;
    }
}

//如果当前解v比目前最优解ans更优，更新ans
bool better(int d) {
    for(int i = d; i >= 0; i--) if(v[i] != ans[i]) {
        return ans[i] == -1 || v[i] < ans[i];
    }
    return false;
}

//当前深度为d，分母不能小于from，分数之和恰好为aa/bb
//往一个方向收敛则无需判重
bool dfs(int d, int from, LL aa, LL bb) {
    if(d == maxd) {
        if(bb % aa) return false; //aa/bb必须是埃及分数
        v[d] = bb/aa;
        if(better(d)) memcpy(ans, v, sizeof(LL) * (d+1));
        return true;
    }

    bool ok = false;
    from = max(from, get_first(aa, bb)); //枚举的起点
    for(int i = from; ; i++) {
        //剪枝：如果剩下的maxd+1-d个分数全部都是1/i，加起来仍然不超过aa/bb，则无解
        if(bb * (maxd+1-d) <= i * aa) break;
        v[d] = i;
        //计算aa/bb - 1/i，设结果为a2/b2
        LL b2 = bb*i;
        LL a2 = aa*i - bb;
        LL g = gcd(a2, b2); //以便约分
        if(dfs(d+1, i+1, a2/g, b2/g)) ok = true;
    }
    return ok;
}

int main(){
    int maxd;
    int a,b;
    cin>>a>>b;

    int ok = 0;
    for(maxd = 1; ; maxd++) {
        memset(ans, -1, sizeof(ans));
        if(dfs(0, get_first(a, b), a, b)) { ok = 1; break; }
    }

    for(int i = 0; i<maxn; ++i){
        if(v[i]){
            cout<<v[i]<<" ";
        }else break;
    }
    return 0;
}