SRM 537 div2

This dish mistaken the time and failed to catch up with t_t. This question does not seem very simple. Today I have made a general introduction.

250pt: Water question

550pt: Give A, B, X, and ask for Y, so that all values of a * P + B * Q can be obtained through X * p '+ y * Q. Return-1 if there are infinite possibilities. Return-1 when X is divisible by A and B at the same time. Y is enumerated in other cases and Y is satisfied (a-y * Q ') % x = 0 and (B-x * p') % Y = 0.

View code

 1 #include <vector>
 2 #include <list>
 3 #include <map>
 4 #include <set>
 5 #include <queue>
 6 #include <deque>
 7 #include <stack>
 8 #include <bitset>
 9 #include <algorithm>
10 #include <functional>
11 #include <numeric>
12 #include <utility>
13 #include <sstream>
14 #include <iostream>
15 #include <iomanip>
16 #include <cstdio>
17 #include <cmath>
18 #include <cstdlib>
19 #include <ctime>
20 
21 using namespace std;
22 
23 
24 class KingXNewCurrency {
25 public:
26     int howMany(int A, int B, int X) {
27         int cnt = 0, f;
28         int y, k;
29         
30         if(!(A%X) && !(B%X))    return -1;
31         int M = max(A, B);
32         for(y = 1; y <= M; ++y) {
33             f = 0;
34             for(k = 0; k <= A; k += y) {
35                 if((A - k)%X == 0) {
36                     f++; break;
37                 }
38             }
39             for(k = 0; k <= B; k += X) {
40                 if((B-k)%y == 0) {
41                     f++; break;
42                 }
43             }
44             if(f == 2)    cnt++;
45         }
46         return cnt;
47     }
48 };
49 
50 
51 <%:testing-code%>
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925pt:

I did not understand the ideas of the experts.

03/22 supplement:

925pt:

I read polla's solution report... I want to understand.

The probability of a cake with a value of-1 being absorbed is 1. If a sequence a-> B-> C-> D exists, it can be seen that a, B, C has a total of 33 cases, but only one case is that D can be absorbed. Therefore, the probability that D is absorbed is 1/a 33.

The final result can accumulate the probability of all elements.

#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;


class PrinceXToastbook {
public:
double eat(vector <int> pre) {
double res = 0, tmp;
int i, j, r, m;
int l = pre.size();
for(i = 0; i < l; ++i) {
            r = i; m = 1;
while(m < 55) {
                r = pre[r];
if(r == -1)    break;
                ++m;
            }
            tmp = 1.0;
for(j = 1; j <= m; ++j) {
                tmp *= j;
            }
            res += 1.0/tmp;
        }
return res;
    }
};


<%:testing-code%>
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