Lightoj 1307-counting Triangles (two minutes)

Title Link: http://lightoj.com/volume_showproblem.php?problem=1307

1307-counting Triangles

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Time Limit:2 second (s)

Memory limit:32 MB

You is given N sticks having distinct lengths; You have a to form some triangles using the sticks. A Triangle is valid if it is positive. Your task is to find the number of ways you can form a valid triangle using the sticks. Input

Input starts with an integer T (≤10), denoting the number of test cases.

Each case is starts with a line containing an integer N (3≤n≤2000). The next line contains N integers denoting the lengths of the sticks. You can assume the lengths is distinct and each length lies in the range [1, 109]. Output

For each case, print the case number and the total number of ways a valid triangle can is formed.

Sample Input	Output for Sample Input
3 5 3 12 5) 4 9 6 1 2 3 4 5 6 4 100 211 212 121	Case 1:3 Case 2:7 Case 3:4

Problem Setter:jane ALAM JAN

Developed and maintained by JANE ALAM JAN

copyright©2012 Lightoj, Jane Alam Jan

Title: Give the length of the triangle, the number of triangles

Topic Analysis: To all sides by the order from small to large, judging is not a triangle, meet a[i] + a[j] > A[k] (i < J < K) can (self-proof,,), and then the range of two k, to seek and

The code is as follows:

Two points (time:412 MS):

#include <iostream> #include <algorithm> #include <map> #include <stack> #include <queue> #include <vector> #include <set> #include <string> #include <cstdio> #include <cstring> #
include<cctype> #include <cmath> #define N 10009 using namespace std;
const int inf = 1E9;
const INT mod = 1<<30;
Const double EPS = 1e-8;
Const double PI = ACOs (-1.0);
typedef long Long LL;

int a[n];
    int main () {int T, n, cnt = 0, I, J;
    scanf ("%d", &t);
        while (t--) {scanf ("%d", &n);
        for (i = 1; I <= n; i++) scanf ("%d", &a[i]);
        Sort (A + 1, a + n + 1);
        int ans = 0, L, R, MD, now; for (i = 1; I <= N, i++) {for (j = i + 1; j <= N; j + +) {L = j; r = N
                ;
                    while (L <= r) {MD = (L + r) >> 1;
           if (A[i] + a[j] > A[MD]) now = MD, L = MD + 1;         else R = md-1;
            } ans + = now-j;
    }} printf ("Case%d:%d\n", ++cnt, ans);
} return 0; }

Non-dichotomy (time:88 MS):

#include <iostream> #include <algorithm> #include <map> #include <stack> #include <queue> #include <vector> #include <set> #include <string> #include <cstdio> #include <cstring> #
include<cctype> #include <cmath> #define N 10009 using namespace std;
const int inf = 1E9;
const INT mod = 1<<30;
Const double EPS = 1e-8;
Const double PI = ACOs (-1.0);
typedef long Long LL;

int a[n];
    int main () {int T, n, cnt = 0, I, J;
    scanf ("%d", &t);
        while (t--) {scanf ("%d", &n);
        for (i = 1; I <= n; i++) scanf ("%d", &a[i]);
        Sort (A + 1, a + n + 1);
        int ans = 0, k;
            for (i = 1; I <= n; i++) {k = i;//must be written so that the second loop will be LTE for (j = i + 1; j <= N; j + +)
                {while (k <= n && a[i] + a[j] > A[k]) k++;
            Ans = ans + k-j-1;
  }} printf ("Case%d:%d\n", ++cnt, ans);  } return 0; }