HDU 5101 Select

  • 2018-08-16
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Description:

One day, Dudu, the most clever boy, heard of ACM/ICPC, which is a very interesting game. He wants to take part in the game. But as we all know, you can’t get good result without teammates.
So, he needs to select two classmates as his teammates.
In this game, the IQ is very important, if you have low IQ you will WanTuo. Dudu’s IQ is a given number k. We use an integer v[i] to represent the IQ of the ith classmate.
The sum of new two teammates’ IQ must more than Dudu’s IQ.
For some reason, Dudu don’t want the two teammates comes from the same class.
Now, give you the status of classes, can you tell Dudu how many ways there are.

Input:

There is a number T shows there are T test cases below. (T≤20)
For each test case , the first line contains two integers, n and k, which means the number of class and the IQ of Dudu. n ( 0≤n≤1000 ), k( 0≤k\<231 ).
Then, there are n classes below, for each class, the first line contains an integer m, which means the number of the classmates in this class, and for next m lines, each line contains an integer v[i], which means there is a person whose iq is v[i] in this class. m( 0≤m≤100 ), v[i]( 0≤v[i]\<231 )

Output:

For each test case, output a single integer.

Sample Input:

1
3 1
1 2
1 2
2 1 1

Sample Output:

5

题目链接

统计所有不同组两数之和大于K的数量。

再输入时对同组数值进行升序排序,之后通过枚举二分统计此组内两数之和大于K的数量,最后对所有数值进行升序排序通过枚举二分统计两数之和大于K的数量,它们之差即为不同组两数之和大于K的数量,此题可以用STL的upper_bound减少代码量。

AC代码:

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