//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//# include <x86intrin.h>
# include <bits/stdc++.h>
#include "koala.h"
# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define _USE_MATH_DEFINES_
#define ll long long
#define ld long double
#define Accepted 0
#define pb push_back
#define mp make_pair
#define sz(x) (int)(x.size())
#define every(x) x.begin(),x.end()
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define For(i,x,y) for (ll i = x; i <= y; i ++)
#define FOr(i,x,y) for (ll i = x; i >= y; i --)
#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
// ROAD to... Red
inline void Input_Output () {
//freopen(".in", "r", stdin);
//freopen(".out", "w", stdout);
}
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
void playRound(int *B, int *R);
int minValue(int N, int W) {
int n = N;
// TODO: Implement Subtask 1 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
int b[N], r[N];
for (int i = 0; i < n; i++)
b[i] = 0;
b[0] = 1;
playRound(b, r);
for (int i = 0; i < N; i++)
if (r[i] == 0)
return i;
return 0;
}
int maxValue(int N, int W) {
// TODO: Implement Subtask 2 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
int n = N;
int u[N] = {};
int b[N], r[N];
for (int i = 0; i < N; i ++)
u[i] = 0;
while (n > 1){
int cost = min(12, W / n);
for (int i = 0; i < N; i++){
if (!u[i])
b[i] = cost;
else
b[i] = 0;
}
playRound(b, r);
for (int i = 0; i < N; i++)
if (!u[i] && r[i] <= b[i])
u[i] = 1;
n = 0;
for (int i = 0; i < N; i++)
if (u[i] == 0)
n ++;
}
for (int i = 0; i < N; i++)
if (!u[i])
return i;
return 0;
}
int greaterValue(int N, int W) {
// TODO: Implement Subtask 3 solution here.
// You may leave this function unmodified if you are not attempting this
// subtask.
int b[N], R[N];
int n = N;
int w = W;
int l = 0, r = 14;
while (r - l > 1) {
int md = (l + r) >> 1;
for (int i = 0; i < n; i++)
b[i] = w / n - 1;
b[0] = b[1] = md * (w / n);
playRound(b, R);
bool g1 = (R[0] > b[0]);
bool g2 = (R[1] > b[1]);
if (g1 == g2){
if (g1 == 1) l = md;
else r = md;
}
else return g2;
}
return 0;
}
void allValues(int N, int W, int *P) {
if (W == 2*N) {
// TODO: Implement Subtask 4 solution here.
// You may leave this block unmodified if you are not attempting this
// subtask.
} else {
// TODO: Implement Subtask 5 solution here.
// You may leave this block unmodified if you are not attempting this
// subtask.
}
}
/*
static int N, W;
static int P[105];
static int maxQueries = 3200;
static int numQueries;
static void runGame(int F);
static void grader();
int main() {
grader();
return 0;
}
void playRound(int *B, int *R) {
int i, j;
int S = 0;
for (i=0;i<N;++i) {
if ( !(B[i] >= 0 && B[i] <= W) ) {
printf("Invalid query.\n");
exit(0);
}
S += B[i];
}
if (S > W) {
printf("Invalid query.\n");
exit(0);
}
numQueries++;
if (numQueries > maxQueries) {
printf("Too many queries.\n");
exit(0);
}
int cache[2][205];
int num[2][205];
char taken[105][205];
for (i=0;i<205;++i) {
cache[1][i] = 0;
num[1][i] = 0;
}
for (i=0;i<N;++i) {
int v = B[i]+1;
int ii = i&1;
int o = ii^1;
for (j=0;j<=W;++j) {
cache[ii][j] = cache[o][j];
num[ii][j] = num[o][j];
taken[i][j] = 0;
}
for (j=W;j>=v;--j) {
int h = cache[o][j-v] + P[i];
int hn = num[o][j-v] + 1;
if (h > cache[ii][j] || (h == cache[ii][j] && hn > num[ii][j])) {
cache[ii][j] = h;
num[ii][j] = hn;
taken[i][j] = 1;
} else {
taken[i][j] = 0;
}
}
}
int cur = W;
for (i=N-1;i>=0;--i) {
R[i] = taken[i][cur] ? (B[i] + 1) : 0;
cur -= R[i];
}
}
static void runGame(int F) {
int i;
scanf("%d %d",&N,&W);
for (i=0;i<N;++i) {
scanf("%d",&P[i]);
}
numQueries = 0;
if (F == 1) {
printf("%d\n", minValue(N, W));
} else if (F == 2) {
printf("%d\n", maxValue(N, W));
} else if (F == 3) {
printf("%d\n", greaterValue(N, W));
} else if (F == 4) {
int userP[105];
allValues(N, W, userP);
for (i=0;i<N;i++) {
printf("%d ",userP[i]);
}
printf("\n");
}
printf("Made %d calls to playRound.\n", numQueries);
}
static void grader() {
int i;
int F, G;
scanf("%d %d",&F,&G);
for (i=0;i<G;i++) {
runGame(F);
}
}
// B...a*/
Compilation message
koala.cpp: In function 'int maxValue(int, int)':
koala.cpp:66:5: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
for (int i = 0; i < N; i ++)
^~~
koala.cpp:69:2: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
while (n > 1){
^~~~~
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
10 ms |
248 KB |
Output is correct |
| 2 |
Correct |
10 ms |
248 KB |
Output is correct |
| 3 |
Correct |
10 ms |
248 KB |
Output is correct |
| 4 |
Correct |
10 ms |
376 KB |
Output is correct |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
22 ms |
248 KB |
Output is correct |
| 2 |
Correct |
22 ms |
376 KB |
Output is correct |
| 3 |
Correct |
23 ms |
376 KB |
Output is correct |
| 4 |
Correct |
22 ms |
376 KB |
Output is correct |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Correct |
68 ms |
248 KB |
Output is correct |
| 2 |
Correct |
76 ms |
248 KB |
Output is correct |
| 3 |
Correct |
70 ms |
252 KB |
Output is correct |
| 4 |
Correct |
67 ms |
332 KB |
Output is correct |
| 5 |
Correct |
66 ms |
248 KB |
Output is correct |
| 6 |
Correct |
66 ms |
248 KB |
Output is correct |
| 7 |
Correct |
67 ms |
352 KB |
Output is correct |
| 8 |
Correct |
70 ms |
248 KB |
Output is correct |
| 9 |
Correct |
69 ms |
248 KB |
Output is correct |
| 10 |
Correct |
79 ms |
252 KB |
Output is correct |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
5 ms |
248 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
5 ms |
248 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
