Ural_1613. For Fans of Statistics(快排+二分)

　　思路就是快排+二分查找。有重复的数字，多加了一点小处理。

My Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 70005;

class node {
public:
    int num;
    int ord;
} a[N];

bool cmp(node a, node b){
    if(a.num == b.num)    return a.ord < b.ord;
    return a.num < b.num;
}

int bin_search(int t, int n) {
    int l, r, mid;
    l = 1; r = n;
    while(l <= r) {
        mid = (l + r) >> 1;
        if(t == a[mid].num)    return mid;
        else if(t < a[mid].num)    r = mid-1;
        else    l = mid+1;
    }
    return 0;
}

int main() {
    //freopen("data.in", "r", stdin);

    int n, i, q, l, r, t, d, flag;
    while(~scanf("%d", &n)) {
        for(i = 1; i <= n; i++) {
            scanf("%d", &a[i].num);
            a[i].ord = i;
        }
        stable_sort(a+1, a+n+1, cmp);
        scanf("%d", &q);
        while(q--) {
            scanf("%d%d%d", &l, &r, &t);
            i = bin_search(t, n);
            d = i; flag = 0;
            //避免重复的数字找不到的情况
            while(t == a[d].num && d > 0) {
                if(a[d].ord >= l && a[d].ord <= r) {
                    printf("1"); flag = 1; break;
                } else d--;
            }
            if(flag)    continue;
            d = i;
            while(t == a[d].num && d <= n) {
                if(a[d].ord >= l && a[d].ord <= r) {
                    printf("1"); flag = 1; break;
                }
                else d++;
            }
            if(flag)    continue;
            printf("0");
        }
        printf("\n");
    }
    return 0;
}