题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=1806
题意：给1~n个单调不递减的数，求q次区间出现次数最多的那个数的次数。
由于单调不递减，是有一个顺序的。则可以考虑记录一个b(i)=第i个数字在前i个里出现了i次，一定是连续的。这样把它丢进st表里，query(l,r)表示的是区间l,r内出现最多的那个数的次数。但是有个问题，那就是最左端有可能不是从１开始的，则要先处理最左端，找到最左的第一个１的位置p，然后用p-l则就是那个数字出现的次数。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &(a)) 43 #define Rll(a) scanf("%lld", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onenum(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long Uint; 62 typedef pair<int, int> pii; 63 typedef pair<LL, LL> pLL; 64 typedef pair<string, LL> psi; 65 typedef map<string, LL> msi; 66 typedef vector<LL> vi; 67 typedef vector<LL> vl; 68 typedef vector<vl> vvl; 69 typedef vector<bool> vb; 70 71 const int maxn = 100100; 72 int n, q; 73 int a[maxn]; 74 int b[maxn]; 75 int dp[maxn][20]; 76 77 void st() { 78 for(int i = 1; i <= n; i++) dp[i][0] = b[i]; 79 int k = int(log(n+1.0)/log(2.0)); 80 for(int j = 1; j <= k; j++) { 81 for(int i = 1; i + (1 << j) - 1 <= n; i++) { 82 dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]); 83 } 84 } 85 } 86 87 int query(int l, int r) { 88 int j = int(log(r-l+1.0)/log(2.0)); 89 return max(dp[l][j], dp[r-(1<<j)+1][j]); 90 } 91 92 int main() { 93 // FRead(); 94 int l, r; 95 while(~Rint(n) && n) { 96 Rint(q); 97 For(i, 1, n+1) { 98 Rint(a[i]); 99 if(i == 1) { 100 b[i] = 1; 101 continue; 102 } 103 if(a[i] == a[i-1]) b[i] = b[i-1] + 1; 104 else b[i] = 1; 105 } 106 st(); 107 W(q) { 108 scanf("%d%d",&l,&r); 109 int m = l; 110 while(m <= r && a[m] == a[m-1]) m++; 111 int ret = max(query(m, r), m - l); 112 printf("%d ", ret); 113 } 114 } 115 RT 0; 116 }