Google Kickstart 2018 Round A题解 | Calvin's Marbles



Google Kickstart 2018 Round A题解 | Calvin's Marbles

这道题蛮简单的，就是要求离N最近的都是数位上都是偶数的数，所以可以向上或者向下找。简单的说就是最高位++或者--，然后剩下来通通改成8或者0。不过我们需要额外考虑一下借位和进位的情况，对于向下来说，不存在，因为1下面是0，不需要借位。对于向上来说9上面是10，要进一位，但是我们注意到进的1是奇数，所以我们还要将它变成2，这无疑是比向下的方法去多了，所以我们不需要考虑9的进位。


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