试题描述
Mr. Frog has an integer sequence of length n, which can be denoted as a1,a2,?,an There are m queries.
In the i-th query, you are given two integers li and ri. Consider the subsequence al_i,al_(i+1),al_(i+2),?,ari.
We can denote the positions(the positions according to the original sequence) where an integer appears first in this subsequence as p(i)1,p(i)2,?,p(i)k_i (in ascending order, i.e.,p(i)1<p(i)2<?<p(i)k_i).
Note that ki is the number of different integers in this subsequence. You should output p(i)?ki/2?for the i-th query.
输入
In the first line of input, there is an integer T (T≤2) denoting the number of test cases.
Each test case starts with two integers n (n≤2×105) and m (m≤2×105). There are n integers in the next line, which indicate the integers in the sequence(i.e., a1,a2,?,an,0≤ai≤2×105).
There are two integers li and ri in the following m lines.
However, Mr. Frog thought that this problem was too young too simple so he became angry. He modified each query to l‘i,r‘i(1≤l‘i≤n,1≤r‘i≤n). As a result, the problem became more exciting.
We can denote the answers as ans1,ans2,?,ansm. Note that for each test case ans0=0.
You can get the correct input li,ri from what you read (we denote them as l‘i,r‘i)by the following formula:
li=min{(l‘i+ansi?1) mod n+1,(r‘i+ansi?1) mod n+1}
ri=max{(l‘i+ansi?1) mod n+1,(r‘i+ansi?1) mod n+1}
输出
You should output one single line for each test case.
For each test case, output one line “Case #x: p1,p2,?,pm”, where x is the case number (starting from 1) and p1,p2,?,pm is the answer.
输入示例
2 5 2 3 3 1 5 4 2 2 4 4 5 2 2 5 2 1 2 2 3 2 4
输出示例
Case #1: 3 3 Case #2: 3 1
数据规模及约定
见“输入”
题解
就是这道题再强行套一个二分。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
int read() {
int x = 0, f = 1; char c = Getchar();
while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = Getchar(); }
while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = Getchar(); }
return x * f;
}
#define maxn 200010
#define maxnode 4000010
int ToT, sumv[maxnode], lc[maxnode], rc[maxnode];
void update(int& y, int x, int l, int r, int p) {
sumv[y = ++ToT] = sumv[x] + 1;
if(l == r) return ;
int mid = l + r >> 1; lc[y] = lc[x]; rc[y] = rc[x];
if(p <= mid) update(lc[y], lc[x], l, mid, p);
else update(rc[y], rc[x], mid + 1, r, p);
return ;
}
int query(int o, int l, int r, int qr) {
if(!o) return 0;
if(r <= qr) return sumv[o];
int mid = l + r >> 1, ans = query(lc[o], l, mid, qr);
if(qr > mid) ans += query(rc[o], mid + 1, r, qr);
return ans;
}
int rt[maxn], lstp[maxn], ANS[maxn], cnt;
int len;
char Out[maxn*7];
int main() {
int T = read();
for(int kase = 1; kase <= T; kase++) {
memset(lstp, 0, sizeof(lstp));
memset(sumv, 0, sizeof(sumv));
memset(lc, 0, sizeof(lc));
memset(rc, 0, sizeof(rc));
memset(rt, 0, sizeof(rt));
ToT = 0;
int n = read(), q = read();
for(int i = 1; i <= n; i++) {
int v = read();
update(rt[i], rt[i-1], 0, n, lstp[v]);
lstp[v] = i;
}
cnt = 0;
int lst = 0;
while(q--) {
int ql = (read() + lst) % n + 1, qr = (read() + lst) % n + 1;
if(ql > qr) swap(ql, qr);
int l = ql, r = qr, k = query(rt[qr], 0, n, ql - 1) - query(rt[ql-1], 0, n, ql - 1) + 1 >> 1, lval = query(rt[ql-1], 0, n, ql - 1);
while(l < r) {
int mid = l + r >> 1;
if(query(rt[mid], 0, n, ql - 1) - lval < k)
l = mid + 1;
else r = mid;
}
ANS[++cnt] = lst = l;
}
printf("Case #%d: ", kase);
len = 0;
int num[10], cntn;
for(int i = 1; i <= cnt; i++) {
int tmp = ANS[i];
if(!tmp) Out[len++] = ‘0‘;
cntn = 0; while(tmp) num[++cntn] = tmp % 10, tmp /= 10;
for(int j = cntn; j; j--) Out[len++] = num[j] + ‘0‘;
if(i < cnt) Out[len++] = ‘ ‘;
}
Out[len] = ‘\0‘;
puts(Out);
}
return 0;
}