C++用 -1 mask法達到branchless解Leetcode1513 Number of Substrings With Only 1...

C++用 -1 mask法達到branchless解Leetcode1513  Number of Substrings With Only 1s|Py3 1-liner

其實很容易，想練功，試試一行解，或是把if分岔拿掉

[C++無分岔解請進]

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It's actually quite easy. If you want to practice, try a one-line solution, or remove if branches.

[codes on Leetcode]https://leetcode.com/problems/number-of-substrings-with-only-1s/solutions/7351052/single-loopbeats-100-by-anwendeng-a9u4/

class Solution {
public:
    const long long mod=1e9+7;
    int numSub(string& s) {
        long long ans=0, cnt=0;
        for(char c: s){
            cnt= -(c=='1') & (cnt+1);
            ans+= -(c=='1') & cnt;
        }
        return ans%mod;
    }
};