题目链接：BZOJ - 2724
题目分析
这道题和 BZOJ-2821 作诗 那道题几乎是一样的，就是直接分块，每块大小 sqrt(n) ，然后将数字按照数值为第一关键字，位置为第二关键字排序，方便之后二分查找某个值在某个区间内出现的次数。
预处理出 f[i][j] 即从第 i 块到第 j 块的答案。
对于每个询问，中间的整块直接用预处理出的，两端的 sqrtn 级别的数暴力做，用二分查找它们出现的次数。
每次询问的复杂度是 sqrtn * logn 。
注意：写代码的时候又出现了给 sort 写的 Cmp() 不保证双向一致的错误！！Warning！注意代码中的 Cmp_Num() 函数，即使在不需要第二关键字的情况下，由于 Num 可能会有相同的，也必须加第二关键字以保证比较结果双向一致！
代码
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
inline void Read(int &Num) {
char c; c = getchar();
while (c < '0' || c > '9') c = getchar();
Num = c - '0'; c = getchar();
while (c >= '0' && c <= '9') {
Num = Num * 10 + c - '0';
c = getchar();
}
}
const int MaxN = 40000 + 5, MaxBlk = 200 + 5;
int n, m, BlkSize, TotBlk;
int A[MaxN], TL[MaxN], T[MaxN], Cnt[MaxN], L[MaxBlk], R[MaxBlk], First[MaxN], Last[MaxN];
int f[MaxBlk][MaxBlk], g[MaxBlk][MaxBlk];
struct ES
{
int Pos, Num, v;
} E[MaxN];
inline bool Cmp_Num(ES e1, ES e2) {
if (e1.Num == e2.Num) return e1.Pos < e2.Pos;
return e1.Num < e2.Num;
}
inline bool Cmp_Pos(ES e1, ES e2) {return e1.Pos < e2.Pos;}
int GetNum(int Num, int x, int y) {
if (x > y || x > E[Last[Num]].Pos || y < E[First[Num]].Pos) return 0;
int l, r, mid, p1, p2;
l = First[Num]; r = Last[Num];
while (l <= r) {
mid = (l + r) >> 1;
if (E[mid].Pos >= x) {
p1 = mid;
r = mid - 1;
}
else l = mid + 1;
}
l = First[Num]; r = Last[Num];
while (l <= r) {
mid = (l + r) >> 1;
if (E[mid].Pos <= y) {
p2 = mid;
l = mid + 1;
}
else r = mid - 1;
}
return p2 - p1 + 1;
}
int main()
{
Read(n); Read(m);
for (int i = 1; i <= n; ++i) {
Read(E[i].Num);
E[i].Pos = i;
}
sort(E + 1, E + n + 1, Cmp_Num);
int v_Index = 0;
for (int i = 1; i <= n; ++i) {
if (i == 1 || E[i].Num > E[i - 1].Num) ++v_Index;
E[i].v = v_Index;
TL[v_Index] = E[i].Num;
}
sort(E + 1, E + n + 1, Cmp_Pos);
for (int i = 1; i <= n; ++i) A[i] = E[i].v;
sort(E + 1, E + n + 1, Cmp_Num);
for (int i = 1; i <= n; ++i) {
if (First[E[i].v] == 0) First[E[i].v] = i;
Last[E[i].v] = i;
}
BlkSize = (int)sqrt((double)n);
TotBlk = (n - 1) / BlkSize + 1;
for (int i = 1; i <= TotBlk; ++i) {
L[i] = (i - 1) * BlkSize + 1;
R[i] = i * BlkSize;
}
R[TotBlk] = n;
for (int i = 1; i <= TotBlk; ++i) {
for (int j = 1; j <= n; ++j) Cnt[j] = 0;
f[i][i - 1] = 0; g[i][i - 1] = 0;
for (int j = i; j <= TotBlk; ++j) {
f[i][j] = f[i][j - 1];
g[i][j] = g[i][j - 1];
for (int k = L[j]; k <= R[j]; ++k) {
++Cnt[A[k]];
if (Cnt[A[k]] > f[i][j] || (Cnt[A[k]] == f[i][j] && A[k] < g[i][j])) {
f[i][j] = Cnt[A[k]]; g[i][j] = A[k];
}
}
}
}
memset(Cnt, 0, sizeof(Cnt));
for (int i = 1; i <= n; ++i) T[i] = -1;
int l, r, x, y, Ct, Ans, Cu;
Ans = 0;
for (int i = 1; i <= m; ++i) {
Read(l); Read(r);
l = (l + Ans - 1) % n + 1; r = (r + Ans - 1) % n + 1;
if (l > r) swap(l, r);
x = (l - 1) / BlkSize + 1; if (l != L[x]) ++x;
y = (r - 1) / BlkSize + 1; if (r != R[y]) --y;
if (x > y) {
Ct = 0; Ans = 0;
for (int j = l; j <= r; ++j) {
++Cnt[A[j]];
if (Cnt[A[j]] > Ct || (Cnt[A[j]] == Ct && A[j] < Ans)) {
Ct = Cnt[A[j]]; Ans = A[j];
}
}
for (int j = l; j <= r; ++j) --Cnt[A[j]];
}
else {
Ct = f[x][y]; Ans = g[x][y];
for (int j = l; j < L[x]; ++j) {
++Cnt[A[j]];
if (T[A[j]] == -1) T[A[j]] = GetNum(A[j], L[x], R[y]);
Cu = Cnt[A[j]] + T[A[j]];
if (Cu > Ct || (Cu == Ct && A[j] < Ans)) {
Ct = Cu; Ans = A[j];
}
}
for (int j = r; j > R[y]; --j) {
++Cnt[A[j]];
if (T[A[j]] == -1) T[A[j]] = GetNum(A[j], L[x], R[y]);
Cu = Cnt[A[j]] + T[A[j]];
if (Cu > Ct || (Cu == Ct && A[j] < Ans)) {
Ct = Cu; Ans = A[j];
}
}
for (int j = l; j < L[x]; ++j) {--Cnt[A[j]]; T[A[j]] = -1;}
for (int j = r; j > R[y]; --j) {--Cnt[A[j]]; T[A[j]] = -1;}
}
Ans = TL[Ans];
printf("%d
", Ans);
}
return 0;
}