Problem Description Chinese always have the railway tickets problem because of its' huge amount of passangers and stations. Now goverment need you to develop a new tickets query system.
One train can just take k passangers. And each passanger can just buy one ticket from station a to station b. Each train cannot take more passangers any time. The one who buy the ticket earlier which can be sold will always get the ticket.

Input The input contains servel test cases. The first line is the case number. In each test case:
The first line contains just one number k( 1 ≤ k ≤ 1000 ) and Q( 1 ≤ Q ≤ 100000 )
The following lines, each line contains two integers a and b, ( 1 ≤ a < b ≤ 1000000 ), indicate a query.
Huge Input, scanf recommanded.
Output For each test case, output three lines:
Output the case number in the first line.
If the ith query can be satisfied, output i. i starting from 1. output an blank-space after each number.
Output a blank line after each test case.
Sample Input 1 3 6 1 6 1 6 3 4 1 5 1 2 2 4
Sample Output Case 1: 1 2 3 5
Author Louty (Special Thanks Nick Gu)
Source

題意：

給出K，和Q，有Q次詢問，火車最多同時不能坐超過 K 個乘客，每個乘客都是按次序登車，

若乘客能滿足上火車的條件則輸出編號，不能就不輸出！

PS:

線段樹查詢區間覆蓋次數！

注意：要求的是輸入的編號的每個數字後面都有空格，最後一個數字也不例外！

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lson l , mid , rt << 1
#define rson mid+1 , r , rt << 1 | 1
const int maxn = 1000010;
const int MAXN = 1000000;
int add[maxn<<2];
int sum[maxn<<2];
int ans[100010];
void PushUp(int rt)
{
    sum[rt] = max(sum[rt<<1], sum[rt<<1|1]);
}
void PushDown(int rt)
{
    if(add[rt])
    {
        add[rt<<1] += add[rt];
        add[rt<<1|1] += add[rt];
        sum[rt<<1] += add[rt];
        sum[rt<<1|1] += add[rt];
        add[rt] = 0;
    }
}

void build(int l, int r, int rt)
{
    add[rt] = 0;
    sum[rt] = 0;//開始都是零
    if(l == r)
    {
        return ;
    }
    int mid = (l+r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}

void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l && r<=R)
    {
        add[rt] += c;
        sum[rt] += c;
        return ;
    }
    PushDown(rt);
    int mid = (l + r) >> 1;
    if(L <= mid)
        update(L, R, c, lson);
    if(mid < R)
        update(L, R, c, rson);
    PushUp(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R)
    {
        return sum[rt];
    }
    PushDown(rt);
    int mid = (l+r) >> 1;
    int ret = 0;
    if(L <= mid)
        ret = max(ret,query(L, R, lson));
    if(mid < R)
        ret = max(ret,query(L, R, rson));
    return ret;
}
int main()
{
    int t;
    int K, Q;
    int cas = 0;
    scanf("%d",&t);
    while(t--)
    {
        memset(ans,0,sizeof(ans));
        scanf("%d%d",&K,&Q);
        int a, b;
        build(1, MAXN, 1);
        int l = 0;
        for(int i = 1; i <= Q; i++)
        {
            scanf("%d%d",&a,&b);
            b--;
            int tt = query(a, b, 1, MAXN, 1);
            if(tt < K)
            {
                ans[++l] = i;
                update(a, b, 1, 1, MAXN, 1);
            }
        }
        printf("Case %d:\n",++cas);
        //printf("%d",ans[1]);
        for(int i = 1; i <= l; i++)
        {
            printf("%d ",ans[i]);
        }
        printf("\n\n");
    }
    return 0;
}