Day6

For the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the container is to support the three operations.

Push: Push a given element e to container

Pop: Pop element of a given e from container

Query: Given two elements a and k, query the kth larger number which greater than a in container;

Although Kiki is very intelligent, she can not think of how to do it, can you help her to solve this problem?

InputInput some groups of test data ,each test data the first number is an integer m (1 <= m <100000), means that the number of operation to do. The next m lines, each line will be an integer p at the beginning, p which has three values:
If p is 0, then there will be an integer e (0 <e <100000), means press element e into Container.

If p is 1, then there will be an integer e (0 <e <100000), indicated that delete the element e from the container  

If p is 2, then there will be two integers a and k (0 <a <100000, 0 <k <10000),means the inquiries, the element is greater than a, and the k-th larger number.
OutputFor each deletion, if you want to delete the element which does not exist, the output "No Elment!". For each query, output the suitable answers in line .if the number does not exist, the output "Not Find!".Sample Input

5
0 5
1 2
0 6
2 3 2
2 8 1
7
0 2
0 2
0 4
2 1 1
2 1 2
2 1 3
2 1 4

Sample Output

No Elment!
6
Not Find!
2
2
4
Not Find!

思路：最后一个操作是时限的关键，而使用树状数组既能快速判断是否是第k大，也能满足前两种操作，套一个二分就行了(等补了线段树再做一次

using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 1e5+5;
const int INF = 0x3f3f3f3f;

int C[maxm], n;

void init() {
    memset(C, 0, sizeof(C));
}

void add(int pos, int val) {
    for(; pos <= maxm; pos += lowbit(pos))
        C[pos] += val;
}

int getsum(int pos) {
    int ret = 0;
    for(; pos; pos -= lowbit(pos))
        ret += C[pos];
    return ret;
}

int biSearch(int a, int k) {
    int l = 1, r = 99999, mid, ans = -1;
    while(l <= r) {
        mid = (l + r) >> 1;
        int t1 = getsum(mid), t2 = getsum(a);
        if(getsum(mid) - getsum(a) >= k) {
            ans = mid;
            r = mid - 1;
        } else 
            l = mid + 1;
    }
    return ans;
}

int main() {
    while(scanf("%d", &n) != EOF) {
        init();
        int type, t1, t2, jud;
        for(int i = 0; i < n; ++i) {
            scanf("%d", &type);
            if(type == 0) {
                scanf("%d", &t1);
                add(t1, 1);
            } else if(type == 1) {
                scanf("%d", &t1);
                if(getsum(t1) - getsum(t1-1) == 0)
                    printf("No Elment!
");
                else
                    add(t1, -1);
            } else if(type == 2) {
                scanf("%d%d", &t1, &t2);
                jud = biSearch(t1, t2);
                if(jud == -1)
                    printf("Not Find!
");
                else
                    printf("%d
", jud);
            }
        }
    }
    return 0;
}