Codeforces round #183 (303c)-Minimum Modular

Two numbers A and B are the same as the remainder M. Obviously (a-B) % m = 0 ..... if M % P = 0... so (a-B) % P = 0 ....

Record all (A [J]-A [I... use a table in C [1000000] to record it... if a number p is the same remainder of the N logarithm .. so C [p] = n...

With the above processing, you can conveniently calculate the MD for the current enumeration... how many pairs will be generated for the same remainder (add all the Integers of MD to C )...

But it does not mean that the number of N-1 needs to be removed. For example...

3, 6, 9, 10 .... (3, 6), (3, 9), (3, 12), (6, 9), (6, 12), (9, 12 )... if the MD of the current test is 3 .. there will be 6 pairs with the remainder of 3 .... in fact, at least three digits are subtracted to ensure that the remainder 3 does not exist...

I have no idea here...

I read other people'sCode... I understood it... it makes sense... but it also feels bold...

In fact, it is not the case that MD is used to calculate the remainder of the same number minus a few numbers .. it is used for pruning .. and then look for brute force attacks (pure brute force .. one by one .. check the number of duplicates .. whether the number of duplicates is not greater than K)

The pruning method is the current MD cool pair> K * (k + 1)/2... find the next MD ....

Why do we not need to use the brute force method to determine the number of K * (k + 1)/2?

Consider an extreme situation .. each number of N is an integer multiple of MD .. obviously, there are N * (n-1)/two same-remainder pairs about MD... however, you only need to remove the n-1 number so that there is no same result...

Think of N-1 as the k... required by the question. Isn't the extreme case that the equivalence pair of MD is (k + 1) * k/2? When the same-remainder pair of MD is> (k + 1) * k/2, delete k + 1 in any case to make the same-remainder pair of MD appear...

This pruning is quite exciting .. although each MD that has passed the same remainder pair <= (k + 1) * k/2 verification must undergo a brute force lookup and judgment... but the time is completely OK...

Program:

 

# Include <iostream> # include <stdio. h> # include <cmath> # include <string. h >#include <algorithm> # include <queue> # include <stack> # define ll long # define oo 1000000000 using namespace STD; int A [5005], c [1000005], n; bool used [1000005]; int main () {int I, J, K, M; while (~ Scanf ("% d", & N, & K) {for (I = 1; I <= N; I ++) scanf ("% d ", & A [I]); sort (a + 1, A + 1 + n); memset (C, 0, sizeof (c); for (I = 1; I <n; I ++) for (j = I; j <= N; j ++) C [A [J]-A [I] ++; for (I = 1; I <= A [n] + 1; I ++) {m = 0; For (j = I; j <= 1000000; J + = I) {M + = C [J]; If (M> K * (k + 1)/2) goto a;} memset (used, false, sizeof (used); M = 0; For (j = 1; j <= N; j ++) {If (used [A [J] % I]) M ++; used [A [J] % I] = true; If (M> K) goto a;} break; :;} printf ("% d \ n", I);} return 0 ;}