Codeforces 602B Approximating a Constant Range（RMQ）

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 100005;
const int maxm = 20;

int N, A[maxn], mx[maxn][maxm+5], mi[maxn][maxm+5];

int query(int l, int r) {
int k = 0;
while ((1<<(k+1)) <= r - l + 1)
k++;
return max(mx[l][k], mx[r-(1<<k)+1][k]) - min(mi[l][k], mi[r-(1<<k)+1][k]);
}

void init () {
scanf("%d", &N);
for (int i = 1; i <= N; i++) scanf("%d", &A[i]);

for (int i = 1; i <= N; i++)
mx[i][0] = mi[i][0] = A[i];

for (int j = 1; (1<<j) <= N; j++) {
for (int i = 1; i + (1<<j) - 1 <= N; i++) {
mi[i][j] = min(mi[i][j-1], mi[i + (1<<(j-1))][j-1]);
mx[i][j] = max(mx[i][j-1], mx[i + (1<<(j-1))][j-1]);
}
}
}

int main () {
init();

int l = 1, r = 1, ans = 1;
while (l <= N) {
while (r <= N && query(l, r) <= 1) r++;
ans = max(r - l, ans);
l++;
}
printf("%d\n", ans);
return 0;
}