LA UVaLive 7371 Triangle (水题，判矩形)

题意：给定两个三角形，问你能不能拼成矩形。

析：很明显，要想是矩形，必须是四个角是直角，那么三角形必须是直角三角形，然后就是只能斜边相对，然后呢？就没了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <functional>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <deque>
#include <map>
#include <cctype>
#include <stack>
#include <sstream>
using namespace std ;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
int a[5], b[5];

int main(){
    while(scanf("%d", &a[0]) == 1){
        for(int i = 1; i < 3; ++i)  scanf("%d", &a[i]);
        for(int i = 0; i < 3; ++i)  scanf("%d", &b[i]);
        sort(a, a+3);  sort(b, b+3);
        if(a[0]*a[0] + a[1]*a[1] == a[2]*a[2] && b[0]*b[0] + b[1]*b[1] == b[2]*b[2]){
            bool ok = true;
            for(int i = 0; i < 3; ++i)
                if(a[i] != b[i]) ok = false;
            if(ok)  puts("YES");
            else puts("NO");
        }
        else puts("NO");
    }
    return 0;
}