Codeforces #275 div2

Link: codeforce #275 div2

A. Counterexample

Given the left and right intervals [l, r], calculate the three numbers a, B, and c in the interval so that a interacts with B,

B and c are mutually qualitative, but a and c are not mutually qualitative. If they do not exist, output-1

Analysis: find the continuous even parity sequence.

# Include <stdio. h> int main () {_ int64 l, r; scanf ("% I64d % I64d", & l, & r); if (l % 2) l ++; if (r-l <2) printf ("-1 \ n"); else printf ("% I64d % I64d % I64d", l, l + 1, l + 2); return 0 ;}

B. Friends and Presents

Construct two sequences. The first sequence must have the number of cnt1 and cannot have a multiple of x,

The second sequence must have the number of cnt2 and cannot have a multiple of y. The two sequences cannot have the same number,

The minimum value of the maximum numbers in the two sequences must be obtained.

Analysis: m = num-num/x, which is the number of numbers from 1 to num that do not contain multiples of x

N = num-num/y, which is the number of numbers from 1 to num that do not contain multiples of x

Num/(x * y) is the number of both x and y.

Therefore, to meet the requirements of m> = cnt1 and n> = cnt2

Because the two sequences cannot have the same number, cnt1 + cnt2 <= num-num/(x * y)

Then, query the minimum value of num in binary mode.

# Include <stdio. h> int main () {_ int64 x, y, cnt1, cnt2, m, n; _ int64 l, r, mid; scanf ("% I64d % I64d % I64d % I64d", & cnt1, & cnt2, & x, & y); l = 1; r = 1e12; while (l <r) {mid = (l + r)/2; m = mid-mid/x; n = mid-mid/y; if (m> = cnt1 & n> = cnt2 & mid-mid/(x * y)> = cnt1 + cnt2) r = mid; else l = mid + 1;} printf ("% I64d \ n", r); return 0 ;}

 

C. Diverse Permutation

Evaluate a series of numbers of n containing 1-n. The number of the absolute values of the two adjacent elements must be k.

Analysis: n numbers have a total of n-1 differences. To ensure that the absolute values of k differences are different,

Then there is the same difference between the n-k-1, you can first output the number of n-k between [1, n-k] in order,

Output the minimum and maximum values of the remaining number in sequence until the number of n is input.

# Include <stdio. h> int main () {int n, k, I, j, num; scanf ("% d", & n, & k); num = n-k-1; for (I = 1; I <= num; I ++) printf ("% d", I); j = n; while (num <n) {printf ("% d", I ++); if (num! = N) printf (""); num ++; if (num = n) break; printf ("% d", j --); if (num! = N) printf (""); num ++;} return 0 ;}

 

Codeforces #275 div2