There are n kids and they want to know who can jump the farthest. For each kid, he can jump three times and the distance he jumps is maximum distance amount all the three jump. For example, if the distance of each jump is (10, 30, 20), then the farthest distance he can jump is 30. Given the distance for each jump of the kids, you should find the rank of each kid.

There are multiple test cases. The first line of input contains an integer T (1≤T≤100), indicating the number of test cases. For each test case: The first line contains an integer n (2≤n≤3), indicating the number of kids. For the next n lines, each line contains three integers ai,bi and ci (1≤ai,bi,ci,≤300), indicating the distance for each jump of the i-th kid. It‘s guaranteed that the final rank of each kid won‘t be the same (ie. the farthest distance each kid can jump won‘t be the same).

For each test case, you should output a single line contain n integers, separated by one space. The i-th integer indicating the rank of i-th kid.

2

3

10 10 10

10 20 30

10 10 20

2

3 4 1

1 2 1

3 1 2

1 2

Hint

For the first case, the farthest distance each kid can jump is 10, 30 and 20. So the rank is 3, 1, 2.

有n小孩在比赛跳远，看谁跳的最远。每个小孩可以跳3次，这个小孩的成绩就是三次距离里面的最大值。例如，一个小孩跳3次的距离分别时10, 30和20，那么这个小孩的成绩就是30。给出每个孩子三次跳的距离，问最终每个孩子的排名是多少。

输入文件的第一行有一个整数T (1≤T≤100)，表示测试数据的组数。对于每组测试数据：第一行包括一个整数n (2≤n≤3), 表示孩子的数目. 接下来n行, 每行包含三个整数ai,bi 和 ci (1≤ai,bi,ci,≤300), 表示第i个小孩每次的跳的距离。输入数据保证每个孩子的成绩互不相同。

对于每组数据，输出一行包含n个整数，用一个空格隔开。第i个数表示第i个小孩的最终名次。

2

3

10 10 10

10 20 30

10 10 20

2

3 4 1

1 2 1

3 1 2

1 2

提示：

对于第一组数据，3个孩子的成绩分别时10, 20和30。因此他们最终排名依次是3, 1和2.

想复杂了，第一次做这种比赛；其实就是把最好的成绩排个顺序，就可以了！

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

struct node
{
    int id;
    int mlen;
}a[105];
int cmp(node x,node y)
{
    return x.mlen>y.mlen;
}
int ss[105];
int main()
{
    int T;
    int ans,i,j,n;
    int x,y,z;
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            int tmp=max(x,y);
            tmp=max(tmp,z);
            a[i].id=i;
            a[i].mlen=tmp;
        }
        sort(a+1,a+n+1,cmp);
        for(i=1;i<=n;i++)
            ss[a[i].id]=i;
        printf("%d",ss[1]);
        for(j=2;j<=n;j++)
            printf(" %d",ss[j]);
        printf("\n");

    }
    return 0;
}

Bestland has a very long road and there are n bus station along the road, which are numbered 1 to n from left to right. There are m persons wanting to take the bus to some other station. You task is to find the time needed for each person. Note: All the other information you need is below. Please read the statment carefully.

There are multiple test cases. The first line of input contains an integer T (1≤T≤60), indicating the number of test cases. For each test case: The first line contains two integers n and m (2≤n,m≤105), indicating the number of bus stations and number of people. In the next line, there are n−1 integers, d1,d2,…,dn−1 (1≤di≤109). The i-th integer means the distance between bus station i and i+1 is di (1≤i<n). In the next m lines, each contains two integers xi and yi (1≤xi,yi≤n,xi≠yi), which means i-th person is in bus station xi and wants goto bus station yi. (1≤i≤m)

What else you should know is that for the i-th person, the bus starts at bus station ((i−1) mod n)+1 and drives to right. When the bus arrives at station n, it will turn around and drive from right to left. Similarly, When the bus arrives at station 1, it will turn around and drive from left to right. You can assume that the bus drives one meter per second. And you should only consider the time that the bus drives and ignore the others.

For each person, you should output one integer which is the minimum time needed before arriving bus station yi.

1

7 3

2 3 4 3 4 5

1 7

4 5

5 4

21

10

28

Hint

For the first person, the bus starts at bus station 1, and the person takes in bus at time 0. After 21 seconds, the bus arrives at bus station 7. So the time needed is 21 seconds. For the second person, the bus starts at bus station 2. After 7 seconds, the bus arrives at bus station 4 and the person takes in the bus. After 3 seconds, the bus arrives at bus station 5. So the time needed is 10 seconds. For the third person, the bus starts at bus station 3. After 7 seconds, the bus arrives at bus station 5 and the person takes in the bus. After 9 seconds, the bus arrives at bus station 7 and the bus turns around. After 12 seconds, the bus arrives at bus station 4. So the time needed is 28 seconds.

Bestland有一条非常长的马路，马路上设有n个公交汽车站。公交汽车站从左到右标号为1到n。有m个人想要乘公交。你的任务是找出每个人到终点为止所需要的时间。注意：你需要用来解决这道题目的信息在Input里面，请仔细阅读。

输入的第一行包含一个整数T (1≤T≤60)，表示测试数据的组数。对于每组测试数据：第一行包含两个整数n和m (2≤n,m≤105)，表示公交车站的数目和乘客的数目。 接下来一行包含n−1个整数, d1,d2,…,dn−1 (1≤di≤109).  di表示第i个公交站和第i+1个公交站之间的距离。在接下来的m行, 每行包含两个整数xi和yi (1≤xi,yi≤n,xi≠yi), 表示第i个人时刻0的时候在第xi个公交站并且想要到第yi个公交站去。(1≤i≤m)

对于第i个人, 公交车在第((i−1) mod n)+1个公交站点在时刻0的时候，并且公交一开始往右开。公交到达站点n的时候会立刻转向往左开，同样当公交到达站点1的时候也会立刻转向往右开。你可以认为公交每秒只开一个单位距离，你只需要考虑公交开的时间。

对于每个人，输出到达yi个公交站点需要的最少时间。

1

7 3

2 3 4 3 4 5

1 7

4 5

5 4

21

10

28

提示：

对于第一个人, 公交在站点1出发, 然后这个人在时刻0上车。21秒之后，公交到达站点7。

对于第二个人，公交在站点2出发。7秒之后，公交到达站点4，这个人上车。之后又过了3秒，公交到达站点5.总共用了10秒。

对于第三个人，公交在站点3出发。7秒之后，公交到达站点5，这个人上车。之后过了9秒，公交达到站点7，然后转向开往站点0。之后经过12秒，公交达到站点4。因此总共需要28秒时间。

Alice has a sequence {a1,a2,…,an} with n positive integers. Bob has some positive integer sequences with different size. Alice wants to know the total occurrences of every sequence Bob has in Alice‘s sequence (the occurrences are allowed to overlap).

We say one sequence B occurs in another sequence A if there is a contiguous subsequence of A that is the same as B after scaled by a positive real factor.

For example A={2,4,8,16},B={1,2} then B occurs three times in A. The occurrences are {2,4}, {4,8} and {8,16}. And the factor is 0.5, 0.25 and 0.125.

There are multiple test cases. The first line of input contains an integer T (1≤T≤30), indicating the number of test cases. For each test case:

The first line contains two integer n and m  (1≤n,m≤100000), indicating the size of Alice‘s sequence and the number of sequences Bob has. In the next line, there are n integers, a1,a2,…,an, indicating Alice‘s sequence. In the following m lines, each starts with an integer ki(1≤ki≤300000) - the size of the sequence. Then ki space separated positive integers follow, indicating the sequence.

The total sum of ki is less than or equal to 1000000. Other integers are between 1 and 10000, inclusive.

For each test case, output a single line with a single integer, indicating the total number of occurrences.

2

4 1

2 4 8 16

2 1 2

5 3

2 4 2 4 6

3 1 2 1

1 5

2 16 8

3

7

Hint

For sample 1, please refer to the problem description. For sample 2, {1, 2, 1} occurs only once, {5} occurs five times and {16, 8} occurs only once.

Alice有一个大小为n的正整数序列{a1,a2,…,an}。Bob也有一些正整数序列。Alice想要知道对于Bob的每个序列，他们在Alice的序列中出现的总次数是多少。

对于两个序列A和B，如果A中有一个连续子序列在乘以某个缩放系数之后和B一样的话，那么B在A中出现。

例如，A={2,4,8,16},B={1,2}, 那么B在A中出现了3次。对应A中的子序列为{2,4}, {4,8}和{8,16}，缩放系数分别是 0.5, 0.25和0.125。

输入第一行包含一个整数T (1≤T≤30)表示测试数据组数。对于每组数据：

第一行包含两个整数n和m  (1≤n,m≤100000)，分别表示Alice的序列大小和Bob拥有的序列的个数。接下来一行包含n个整数，a1,a2,…,an，表示Alice的序列。接下来m行，每行的第一个数是ki(1≤ki≤300000)表示这个序列的大小。接下来ki个用空格分开的正整数，表示这个序列。

保证所有ki的和小于等于1000000。其他整数范围介于1和10000之间（包括1和10000）。

对于每组数据，输出一个整数表示出现的总次数。

2

4 1

2 4 8 16

2 1 2

5 3

2 4 2 4 6

3 1 2 1

1 5

2 16 8

3

7

Bob has an array {a1,a2,…,an}, each element is an integer between 1 and n. Bob also has m functions whose domain and codomain are both {1,2,…,n}. Bob and Alice begin to play a game on the array. Alice plays first. For each turn, the player can choose a function f and make every ai (1≤i≤n) become f(ai). For example, the array is {1,1,2,4,5} and f(1)=1,f(2)=3,f(3)=4,f(4)=1,f(5)=2. Then after the operation the array will become {1,1,3,1,2}. If all the element in the array is same, then Alice wins and the game stops. So you can see that Bob‘s goal is to stop Alice. Suppose that both Alice and Bob play optimally. Alice wants to know if she can always win no matter what the array looks like.

There are multiple test cases. The first line of input contains an integer T (1≤T≤200), indicating the number of test cases. For each test case: The first line contains two integers n and m (1≤n,m≤100), indicating the element in the array and the number of functions. In the next m lines, each contains n integers f(1),f(2),…,f(n) (1≤f(i)≤n,1≤i≤n).

For each case, output "YES"(without quotes) if Alice can always win no matter what the array looks like, otherwise output "NO"(without quotes).

2

5 1

1 3 4 1 2

5 1

2 3 4 5 1

YES

NO

Bob有一个数组{a1,a2,…,an}，数组中的每个元素都是介于1到n之间的整数。Bob还有m 个函数，他们的定义域和值域都是集合{1,2,…,n}。 Bob和Alice轮流开始玩游戏，Alice先开始。 对于每一轮，玩家可以选择一个函数f使得数组中每个元素 ai (1≤i≤n)变成f(ai)。例如，一开始数组是{1,1,2,4,5}，有一个函数f(1)=1,f(2)=3,f(3)=4,f(4)=1,f(5)=2。那么经过一次操作，数组变为{1,1,3,1,2}。如果数组中的所有元素都相同（无论当前论是Bob还是Alice），那么Alice胜利游戏结束。然后Bob的目的是阻止Alice胜利。 假设Alice和Bob都足够聪明，每次都采取最优策略。问：无论数组的初始状态是什么，Alice是否都能够必胜？

输入第一行包含一个整数T (1≤T≤200)表示测试数据组数。对于每组测试数据：第一行包含两个整数n和m (1≤n,m≤100)表示数组的大小和函数的个数。接下来m行，每行包含n个整数f(1),f(2),…,f(n) (1≤f(i)≤n,1≤i≤n)。

对于每组数据，如果Alice一定能够必胜，输出YES，否则输出NO。

2

5 1

1 3 4 1 2

5 1

2 3 4 5 1

YES

NO