暫時只做了遞增序列的求解方法,就是將原數列處理成個數的數組,類似於前綴和的東西,對這個頻率數組求區間最大值,
但是僅僅維護區間max還是不夠的,,好比這種情況:
原數列: 1 1 1 2 3 5 5 7 7
頻率數組: 1 2 3 1 1 1 2 1 2
下標: 1 2 3 4 5 6 7 8 9
如果查詢區間3~5,頻率數組分別爲3 1 1這樣,max就是3明顯不對,這時加一種處理,去掉開始的重複部分,就是保證查詢區間的第一個數的頻率數組權值爲1。詳情見代碼:
AC代碼:
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <fstream>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
//#define maxd 1010
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define mc(x, y) memcpy(x, y, sizeof(x))
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,n) for(int i=0;i<(n);i++)
#define repf(i,a,b) for(int i=(a);i<=(b);i++)
#define PI pair<int,int>
//#define mapp make_pair
#define FI first
#define SE second
#define IT iterator
#define PB push_back
#define Times 10
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int > pce;
//#define N 100
const double eps = 1e-10;
const double pi = acos(-1.0);
const ll mod = 1e9+7;
const int inf = 0x3f3f3f3f;
//const ll INF = (ll)1e18+300;
const int maxd = 100500 + 10;
int n;
int Q;
int ac[maxd];
int max_[maxd][20];
int min_[maxd][20];
int pre[maxd];
void RMQ() {
for (int i = 1; i <= n; i++) {
max_[i][0] = min_[i][0] = pre[i];
}
for (int j = 1; j <= 20; j++) {
/*外層循環要從j開始,因爲ST表我們是通過j-1來更新j的,如果是i爲外層循環,第一次我們計算出了:
1, 2, 4, 8,...這些位置,而下一次更新的時候,我們要的是max([1,4], [5, 8])但是我們沒有計算出這兩個區間的最大值,因爲一次層的時候
沒有計算3, 5, 6, 7,...的值,因此必須j爲第一層循環。
*/
for (int i = 1; i <= n; i++) {
if(i + (1<<j) - 1 <= n) {
max_[i][j] = max(max_[i][j-1], max_[i + (1<<(j-1))][j - 1]);
//min_[i][j] = min(min_[i][j-1], min_[i + (1<<(j-1))][j - 1]);
}
}
}
}
int main() {
while(~scanf("%d", &n) && n) {
scanf("%d", &Q);
for (int i = 1; i <= n; i++) {
scanf("%d", &ac[i]);
}
//sort(ac+1, ac+1+n);
for (int i = 1; i <= n; i ++) {
if(i == 1) {
pre[i] = 1;
continue;
}
if(ac[i] == ac[i - 1]) {
pre[i] = pre[i - 1] + 1;
}
else {
pre[i] = 1;
}
}
RMQ();
for (int i = 1; i <= Q; i++) {
int a, b;
scanf("%d%d", &a, &b);
int temp = a;
while(temp <= b && ac[temp] == ac[temp - 1]) {
temp ++;
}
int res = temp - a;
a = temp;
if(a <= b) {
int k = (int)(log(b - a + 1)/log(2.0));
/*查詢操作就像圖片中所示,取超過(r - l)/2的長度,分區間比較,而我們就可以取log2並向上取整.
*/
int max_ans = max(max_[a][k], max_[b - (1<<k) + 1][k]);
//cout << max_ans << " " << res << endl;
//int min_ans = min(min_[a][k], min_[b - (1<<k) + 1][k]);
cout << max(res, max_ans) << endl;
}
else {
cout << res << endl;
}
}
}
}