ACdream OJ 1140 Counting Triangles

　　给出N条边，问这些边能组成多少个不同的三角形，每条边的长度是不同的。首先我们先

将边长按照升序排序，然后枚举两条短的边，看最长边有多少种可能，累加。

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

const int MAXN = 2012;
int T, N;
int a[MAXN];

int b_s( int len)
{
    int left = 0, right = N - 1;
    while( left < right) {
        int mid = ( left + right + 1) >> 1;
        if( a[mid] >= len)
            right = mid - 1;

        else
            left = mid;
    }
    return left;
}

int cal()
{
    int cnt = 0;
    for( int i = 0; i < N; i ++)
        for( int j = i + 1; j < N; j ++) {
            int len = a[j] + a[i];
            int pos = b_s( len);
            cnt += (pos - j);
        }
    return cnt;
}

int main()
{
    scanf( "%d", &T);
    for( int cas = 1; cas <= T; cas ++)
    {
        scanf( "%d", &N);
        for( int i = 0; i < N; i ++)
            scanf( "%d", &a[i]);
        sort( a, a + N);
        int ans = cal();
        printf( "Case %d: %d\n", cas, ans);
    }
    return 0;
}