求 1+2+...+n ，要求不能使用乘除法、for、while、if、else、switch、case等关键字及条件判断语句（A?B:C）。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/qiu-12n-lcof
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class Solution {

    private int multiply(int a, int b) {
        int ans = 0;
        while (b > 0) {
            if ((b & 1) == 1) {
                ans += a;
            }
            a <<= 1;
            b >>= 1;
        }
        return ans;
    }


    /**
     * (n + 1) * n / 2
     */
    public int sumNums(int n) {
        int ans = 0;
        int a = n, b = n + 1;
        boolean inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        inf = (b & 1) == 1 && (ans += a) > 0;
        a <<= 1;
        b >>= 1;
        return ans >> 1;
    }
}