// Find all substrings that % m == l
// Use Horner's algorithm to quickly evaluate sum(a_n*10^n) where
// a_n are the segments' durabilities
int count_segments(int* first, int* last, int m, int l) {
int n = 0, number = 0;
while( first<last ) {
number = number * 10 + *first; // This is Horner's method
if( (number % m)==l ) {
n++;
// If you don't believe - enable this line of output and
// see the numbers matching the combinations of the
//cout << "[" << m << ", " << l << "]: " << number << endl;
}
first++;
}
return n;
}
我想在设置容器后询问第一个盒子是否无法绘制? 因为我的.Review没有画一个盒子......但是.Review1做了!! !! 我想在容器内放两个盒子,这可能吗?