hdu 5101 Select
9979 ワード
タイトル接続
http://acm.hdu.edu.cn/showproblem.php?pid=5101
Select
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. $(T\leq 20)$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\leq n\leq 1000), k\( 0\leq k\leq 2^{31} ).$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\leq m\leq 100 ), v[i]\( 0\leq v[i] < 2^{31} )$
Output
For each test case, output a single integer.
Sample Input
13 11 21 22 1 1
Sample Output
5
構想:すべての数から選択された2つの加算とkより大きい数のシナリオ数-同じ集合で選択された2つの加算とkより大きい数のシナリオ数..
まず並べ替えて、それから2点.の
View Code
http://acm.hdu.edu.cn/showproblem.php?pid=5101
Select
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. $(T\leq 20)$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\leq n\leq 1000), k\( 0\leq k\leq 2^{31} ).$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\leq m\leq 100 ), v[i]\( 0\leq v[i] < 2^{31} )$
Output
For each test case, output a single integer.
Sample Input
13 11 21 22 1 1
Sample Output
5
構想:すべての数から選択された2つの加算とkより大きい数のシナリオ数-同じ集合で選択された2つの加算とkより大きい数のシナリオ数..
まず並べ替えて、それから2点.の
1 #include<algorithm>
2 #include<iostream>
3 #include<cstdlib>
4 #include<cstring>
5 #include<cstdio>
6 #include<vector>
7 #include<map>
8 #include<set>
9 using std::cin;
10 using std::cout;
11 using std::endl;
12 using std::find;
13 using std::sort;
14 using std::set;
15 using std::map;
16 using std::pair;
17 using std::vector;
18 using std::lower_bound;
19 #define pb(e) push_back(e)
20 #define sz(c) (int)(c).size()
21 #define mp(a, b) make_pair(a, b)
22 #define all(c) (c).begin(), (c).end()
23 #define iter(c) decltype((c).begin())
24 #define cls(arr,val) memset(arr,val,sizeof(arr))
25 #define cpresent(c, e) (find(all(c), (e)) != (c).end())
26 #define rep(i, n) for (int i = 0; i < (int)(n); i++)
27 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
28 const int N = 1010;
29 typedef long long ll;
30 int num[N], arr[N * 100], rec[N][102];
31 int main() {
32 #ifdef LOCAL
33 freopen("in.txt", "r", stdin);
34 freopen("out.txt", "w+", stdout);
35 #endif
36 ll ans;
37 int t, n, k, tot;
38 scanf("%d", &t);
39 while (t--) {
40 ans = tot = 0;
41 scanf("%d %d", &n, &k);
42 rep(i, n) {
43 scanf("%d", &num[i]);
44 rep(j, num[i]) scanf("%d", &rec[i][j]), arr[tot++] = rec[i][j];
45 }
46 sort(arr, arr + tot);
47 rep(i, tot) ans += tot - (lower_bound(arr + i, arr + tot, k - arr[i] + 1) - arr);
48 rep(i, n) {
49 sort(rec[i], rec[i] + num[i]);
50 rep(j, num[i]) {
51 ans -= num[i] - (lower_bound(rec[i] + j, rec[i] + num[i], k - rec[i][j] + 1) - rec[i]);
52 }
53 }
54 printf("%lld
", ans);
55 }
56 return 0;
57 }View Code