算法竞赛进阶指南--在单调递增序列a中查找>=x的数中最小的一个(即x或x的后继)
while (l < r) {
int mid = (l + r) / 2;
if (a[mid] >= x) r = mid; else l = mid + 1;
}
while (l < r) {
int mid = (l + r) / 2;
if (a[mid] >= x) r = mid; else l = mid + 1;
}