UVaLive 7371 Triangle (水题，判矩形)

时间：2024-01-11 16:28:20

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;

}

　　

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