#include <iostream>
#include <string>
using namespace std;
int n, k, t;
string input;
int result = -1;
int prefix1[10000];
int prefix2[10000];
int prefix3[10000];
int get1(int a, int b)
{
if (a==0)
{
return prefix1[b];
}
else
{
return prefix1[b] - prefix1[a-1];
}
}
int get2(int a, int b)
{
if (a==0)
{
return prefix2[b];
}
else
{
return prefix2[b] - prefix2[a-1];
}
}
int get3(int a, int b)
{
if (a==0)
{
return prefix3[b];
}
else
{
return prefix3[b] - prefix3[a-1];
}
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n >> k >> t;
cin >> input;
if (input[0] == '1')
{
prefix1[0] = 1;
prefix2[0] = 0;
prefix3[0] = 0;
}
else if (input[0] == '2')
{
prefix1[0] = 0;
prefix2[0] = 1;
prefix3[0] = 0;
}
else if (input[0] == '3')
{
prefix1[0] = 0;
prefix2[0] = 0;
prefix3[0] = 1;
}
for (int i = 0; i < input.size(); i++)
{
prefix1[i] = prefix1[i-1];
prefix2[i] = prefix2[i-1];
prefix3[i] = prefix3[i-1];
if (input[i] == '1')
{
prefix1[i]++;
}
else if (input[i] == '2')
{
prefix2[i]++;
}
else if (input[i] == '3')
{
prefix3[i]++;
}
}
int nie_mozliwe_meatingi = 0;
int ile_trwa_podroz = 0;
int na_ilu_spodkaniach_musze_byc_po_za_biurem = 0;
int ile_jestem_w_biurze = 0;
if(k < get1(0, n-1)) // musi isc do biura
{
for (int i = t - 1; i < n; i++) // i dojazd (ostatnia godzina podrozy z domu do biura)
{
for (int j = i + 1; j < n-t+1; j++) // j odjazd (pierwsza godzina podrozy z biura do domu)
{
nie_mozliwe_meatingi = 0;
nie_mozliwe_meatingi += get1(0, i) + get1(j, n - 1); // na ilu obowiazkowych z biura nie jestem
nie_mozliwe_meatingi += get2(i-t+1, i) + get2(j, j + t - 1); // na ilu zdalnych nie jestem bo w podruzy
ile_trwa_podroz = t*2;
na_ilu_spodkaniach_musze_byc_po_za_biurem = (get2(0, i-t) + get2(j + t, n-1)) - (k - nie_mozliwe_meatingi);
ile_jestem_w_biurze = j - i - 1;
if (nie_mozliwe_meatingi <= k)
{
result = max(result, n - ile_jestem_w_biurze - ile_trwa_podroz - na_ilu_spodkaniach_musze_byc_po_za_biurem );
}
}
}
}
else
{
result = min(k, get1(0, n-1) + get2(0, n-1)) + get3(0, n-1);
}
cout << result << '\n';
return 0;
}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 | #include <iostream> #include <string> using namespace std; int n, k, t; string input; int result = -1; int prefix1[10000]; int prefix2[10000]; int prefix3[10000]; int get1(int a, int b) { if (a==0) { return prefix1[b]; } else { return prefix1[b] - prefix1[a-1]; } } int get2(int a, int b) { if (a==0) { return prefix2[b]; } else { return prefix2[b] - prefix2[a-1]; } } int get3(int a, int b) { if (a==0) { return prefix3[b]; } else { return prefix3[b] - prefix3[a-1]; } } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> n >> k >> t; cin >> input; if (input[0] == '1') { prefix1[0] = 1; prefix2[0] = 0; prefix3[0] = 0; } else if (input[0] == '2') { prefix1[0] = 0; prefix2[0] = 1; prefix3[0] = 0; } else if (input[0] == '3') { prefix1[0] = 0; prefix2[0] = 0; prefix3[0] = 1; } for (int i = 0; i < input.size(); i++) { prefix1[i] = prefix1[i-1]; prefix2[i] = prefix2[i-1]; prefix3[i] = prefix3[i-1]; if (input[i] == '1') { prefix1[i]++; } else if (input[i] == '2') { prefix2[i]++; } else if (input[i] == '3') { prefix3[i]++; } } int nie_mozliwe_meatingi = 0; int ile_trwa_podroz = 0; int na_ilu_spodkaniach_musze_byc_po_za_biurem = 0; int ile_jestem_w_biurze = 0; if(k < get1(0, n-1)) // musi isc do biura { for (int i = t - 1; i < n; i++) // i dojazd (ostatnia godzina podrozy z domu do biura) { for (int j = i + 1; j < n-t+1; j++) // j odjazd (pierwsza godzina podrozy z biura do domu) { nie_mozliwe_meatingi = 0; nie_mozliwe_meatingi += get1(0, i) + get1(j, n - 1); // na ilu obowiazkowych z biura nie jestem nie_mozliwe_meatingi += get2(i-t+1, i) + get2(j, j + t - 1); // na ilu zdalnych nie jestem bo w podruzy ile_trwa_podroz = t*2; na_ilu_spodkaniach_musze_byc_po_za_biurem = (get2(0, i-t) + get2(j + t, n-1)) - (k - nie_mozliwe_meatingi); ile_jestem_w_biurze = j - i - 1; if (nie_mozliwe_meatingi <= k) { result = max(result, n - ile_jestem_w_biurze - ile_trwa_podroz - na_ilu_spodkaniach_musze_byc_po_za_biurem ); } } } } else { result = min(k, get1(0, n-1) + get2(0, n-1)) + get3(0, n-1); } cout << result << '\n'; return 0; } |
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