# 【bzoj2724】[Violet 6]蒲公英

*题目描述：

*输入：

l = (l_0 + x - 1) mod n + 1, r = (r_0 + x - 1) mod n + 1
*输出：

*样例输入：
6 3
1 2 3 2 1 2
1 5
3 6
1 5
*样例输出：
1
2
1
*提示：

n <= 40000, m <= 50000
*题解：

*代码：

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

#ifdef WIN32
#define LL "%I64d"
#else
#define LL "%lld"
#endif

#ifdef CT
#define debug(...) printf(__VA_ARGS__)
#define setfile()
#else
#define debug(...)
#define filename ""
#define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout);
#endif

#define R register
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)
#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))
#define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))
#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)
#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)
char B[1 << 15], *S = B, *T = B;
inline int FastIn()
{
R char ch; R int cnt = 0; R bool minus = 0;
while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;
ch == '-' ? minus = 1 : cnt = ch - '0';
while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';
return minus ? -cnt : cnt;
}
#define maxn 40010
int n, m, block, num, ans, anscnt;
int hash[maxn], a[maxn], id[maxn], cnt[maxn], Ans[210][210], Cnt[210][maxn];
int main()
{
//  setfile();
n = FastIn(), m = FastIn();
block = sqrt(n); num = n / block;
R int tot = n;
for (R int i = 1; i <= n; ++i) a[i] = hash[i] = FastIn(), id[i] = i / block;
std::sort(hash + 1, hash + tot + 1);
tot = std::unique(hash + 1, hash + tot + 1) - hash - 1;
for (R int i = 1; i <= n; ++i)
a[i] = std::lower_bound(hash + 1, hash + tot + 1, a[i]) - hash;
// lisanhua
for (R int i = 0; i <= num; ++i)
for (R int j = 1; j < dmin((i + 1) * block, n + 1); ++j)
++Cnt[i][a[j]];
for (R int i = 0; i < num; ++i)
{
anscnt = 0;
for (R int j = i; j <= num; ++j)
{
for (R int k = j * block; k < dmin((j + 1) * block, n + 1); ++k)
++cnt[a[k]] > anscnt || (cnt[a[k]] == anscnt && a[k] < ans)
? anscnt = cnt[a[k]], ans = a[k]
: 0;
Ans[i][j] = ans;
}
memset(cnt, 0, sizeof(cnt));
}
ans = 0;
for (; m; --m)
{
R int ql = (FastIn() + ans - 1) % n + 1, qr = (FastIn() + ans - 1) % n + 1;
anscnt = 0;
if (ql > qr) std::swap(ql, qr);
if (id[ql] == id[qr])
{
for (R int i = ql; i <= qr; ++i)
++cnt[a[i]] > anscnt || (cnt[a[i]] == anscnt && a[i] < ans)
? anscnt = cnt[a[i]], ans = a[i]
: 0;
for (R int i = ql; i <= qr; ++i) --cnt[a[i]];
}
else
{
R int ls = id[ql] + 1, rs = id[qr] - 1;
ans = Ans[ls][rs]; anscnt = Cnt[rs][ans] - Cnt[ls - 1][ans];
for (R int i = ql; i < ls * block; ++i)
{
R int tmp = ++cnt[a[i]] + Cnt[rs][a[i]] - Cnt[ls - 1][a[i]];
tmp > anscnt || (tmp == anscnt && a[i] < ans)
? anscnt = tmp, ans = a[i]
: 0;
}
for (R int i = (rs + 1) * block; i <= qr; ++i)
{
R int tmp = ++cnt[a[i]] + Cnt[rs][a[i]] - Cnt[ls - 1][a[i]];
tmp > anscnt || (tmp == anscnt && a[i] < ans)
? anscnt = tmp, ans = a[i]
: 0;
}
for (R int i = ql; i < ls * block; ++i) --cnt[a[i]];
for (R int i = (rs + 1) * block; i <= qr; ++i) --cnt[a[i]];
}
printf("%d\n", ans = hash[ans]);
}
return 0;
}
posted @ 2016-05-22 15:59  cot  阅读(176)  评论(0编辑  收藏  举报