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 << endlconst 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;}