http://codeforces.com/problemset/problem/758/C
C. Unfair Poll
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.
Seating in the class looks like a rectangle, where n rows with m pupils in each.
The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, …, the n - 1-st row, the n-th row, the n - 1-st row, …, the 2-nd row, the 1-st row, the 2-nd row, …
The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, …, the m-th pupil.
During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:
If there is only one row in the class, then the teacher always asks children from this row.
Input
The first and the only line contains five integers n, m, k, x and y (1 ≤ n, m ≤ 100, 1 ≤ k ≤ 1018, 1 ≤ x ≤ n, 1 ≤ y ≤ m).
Output
Print three integers:
Examples
Input
1 3 8 1 1
Output
3 2 3
Input
4 2 9 4 2
Output
2 1 1
Input
5 5 25 4 3
Output
1 1 1
Input
100 100 1000000000000000000 100 100
Output
101010101010101 50505050505051 50505050505051
Note
The order of asking pupils in the first test:
The order of asking pupils in the second test:
思路:可以发现排是这样的1,2,3,….n, n-1,…3,2;这样是一个循环,一个循环是aa=(2×n-2)×m
那么除了1,n排是k/aa,其余的为2×k/aa,那么剩下的k%aa,再模拟下就可以了。然后在m[i][j] (模拟后每个位置上问的问题)找maxx,minn,和m[x][y];
1 #include
2 using namespace std;
3 typedef long long LL;
4 LL ma[200][200];
5 int main(void)
6 {
7 LL n,m,k,x,y;
8 scanf("%lld %lld %lld %lld %lld",&n,&m,&k,&x,&y);
9 LL ak = n*m;
10 LL maxx = 0,minn = 0;
11 LL t = 0;
12 if(n == 1)
13 {
14 int xx = 1;
15 int yy = 0;
16 maxx = k/ak;
17 minn = k/ak;
18 if(k%ak)maxx++;
19 LL ac = k%ak;
20 int d = 0;
21 LL ask = minn ;
22 if(y <= ac)
23 ask++;
24 printf("%lld %lld %lld\n",maxx,minn,ask);
25 }
26 else
27 {
28 LL acc = (2*n-2)*m;minn = k/acc;
29 if(n == 2)
30 {
31 maxx = k/acc;
32 }
33 else
34 {
35 maxx = k/acc*(LL)2;
36 }
37 for(int i = 1; i <= n; i++)
38 {
39 if(i == 1||i == n)
40 {
41 for(int j = 1; j <= m; j++)
42 {
43 ma[i][j] = minn;
44 }
45 }
46 else
47 {
48 for(int j = 1; j <= m; j++)
49 ma[i][j] = maxx;
50 }
51 }
52
53 LL low = k%acc;
54 int xx = 1;
55 int yy = 0;
56 int d = 0;
57 while(low)
58 {
59 yy++;
60 low--;
61 ma[xx][yy]++;
62 if(yy == m)
63 {
64 yy = 0;
65 if(d == 0)
66 xx++;
67 if(d == 1)
68 xx--;
69 if(xx == n+1)
70 xx = n-1,d = 1;
71 if(xx == 0)
72 xx = 2,d = 0;
73 }
74 }
75 minn = 1e16;
76 for(int i = 1; i <= n; i++)
77 {
78 for(int j = 1; j <= m; j++)
79 {
80 maxx = max(maxx,ma[i][j]);
81 minn = min(minn,ma[i][j]);
82 }
83 }
84 LL ask = ma[x][y];
85 printf("%lld %lld %lld\n",maxx,minn,ask);
86 }
87 return 0;
88 }
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