Ural 1091 Tmutarakan Exams "the principle of tolerance and repulsion"

Topic Links:

http://acm.timus.ru/problem.aspx?space=1&num=1091

Main topic:

give you two integers k and s, select the number of K in a nonnegative integer equal to S, and the number of K greatest common divisor is greater than 1,

Ask how many groups there are in total. (2 <= K <= S <= 50).

Problem Solving Ideas:

Since 2 <= k <= S <= 50, we can directly enumerate the factorization and find out the number of K selected from each factorization multiple

Number of combinations, adding up is the number of scenarios, but this repeated calculation of a lot of cases.

For example: S = 20,k = 2.

Multiples of 2:2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Multiples of 3:3, 6, 9, 12, 15, 18

Multiples of 5:5, 10, 15, 20

Multiples of 7:7, 14

Multiples of 11:11

......

if they are simply cumulative, most of them are repeated several times. You should remove the extra parts. equivalent to a multiple of 2,

multiples of 3 、... All as a collection of each part of the combination of the case, and then to find the set of the assembly.

if the number of K in the 2-fold set is P (2), the results are as follows:

P (2) +p (3) +...+p (23°c)-P (2*3)-P (2*5) .... (2*11) + P (2*3*5) + ...

Since s maximum number does not exceed 50,k minimum of 2, then 2*29 = > 50, so the mass factor is enumerated to 23.

The aggregation of a set is made by using the principle of tolerance.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace    Std;int prime[22] = {2,3,5,7,11,13,17,19,23,29};int C (int n,int m) {if (n < m)//No error return 0;    m = min (M, n-m);    int ans = 1;        for (int i = 1; I <= m; ++i) {ans *= (n-i+1);    Ans/= i; } return ans;    int K,s;int Solve () {int ans = 0,top = 0;            for (int i = 0; i < ++i) {if (S/prime[i] < K) {top = i;        Break        }} for (int i = 1; i < (1 << top); ++i) {int odd = 0,mult = 1;  for (int j = 0; J < top;                ++J) {if (1 << j) & i) {odd++;            Mult *= Prime[j];        }} if (Odd & 1) ans + = C (S/mult, K);    else ans-= C (S/mult, K);    } if (ans <= 10000) return ans; else return 10000;} int main () {while (~SCANF ("%d%d ", &k,&s)) {printf ("%d\n ", Solve ()); } return 0;}

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Ural 1091 Tmutarakan Exams "the principle of tolerance and repulsion"