洛谷P2036

題目鏈接：洛谷P2036

一看數據範圍，直接想到二進制枚舉子集

int main()
{
    // fre;
    int n;
    cin >> n;
    int ans = INF;
    int a[12], b[12];
    for (int i = 1; i <= n; i++)
        cin >> a[i] >> b[i];
    for (int i = 1; i < (1 << n); i++)
    {
        int s1 = 1;
        int s2 = 0;
        for (int j = 0; j < n; j++)
            if ((1 << j) & i)
                s1 *= a[j + 1], s2 += b[j + 1];
        ans = (ans > fabs(s1 - s2)) ? fabs(s1 - s2) : ans;
    }
    cout << ans << endl;
    return 0;
}

另外也可以DFS，不過我覺得思路和二進制枚舉子集差不多。

int n;
int a[12], b[12];
int ans = INF;
void dfs(int now, int s1, int s2, int ff)
{
    if (now > n + 1)
        return;
    dfs(now + 1, s1 * a[now], s2 + b[now], 1);
    dfs(now + 1, s1, s2, ff);
    if (now == n + 1 && ff)
        ans = min(ans, (int)fabs(s1 - s2));
}
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i] >> b[i];
    dfs(1, 1, 0, 0);
    cout << ans << endl;
    return 0;
}