Educational Codeforces Round 7 C. Not Equal on a Segment（思维）

题意:

给定N,Q≤2×105,N为序列长度,Q次询问
li,ri,xi,查找[li,ri]中≠xi的任意一个下标,无解输出−1

分析:

经典想法了,R[i]:=与a[i]相等的最右的那个下标
R[i]=a[i]==a[i+1] ? R[i+1]:i
询问的时候判断a[l]=x与否,不然就是R[l]+1,看是否在r内
时间复杂度为O(n+q)

由于ai≤106,可以维护106个vector来保存下标
然后二分x这个vector里[li,ri]子集的[l′,r′],如果r′−l′==ri−li,就是−1
不然就再次二分找到那个缺口,就是答案
时间复杂度为O(n+qlog2n)

代码:

//
// Created by TaoSama on 2016-02-10
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, q;
int a[N], R[N];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; ++i) scanf("%d", a + i);
    for(int i = n; i; --i)
        if(a[i] == a[i + 1]) R[i] = R[i + 1];
        else R[i] = i;
    while(q--) {
        int l, r, x; scanf("%d%d%d", &l, &r, &x);
        if(a[l] != x) printf("%d\n", l);
        else {
            if(R[l] + 1 <= r) printf("%d\n", R[l] + 1);
            else puts("-1");
        }
    }
    return 0;
}