Codeforces Beta Round #6 (Div. 2 Only) A. Triangle
2015-02-17 21:23
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题目大意
给出四条边,问是否可以取三条边组成一个三角形,如果不行的话可否组成一个退化的三角形(两边之和等于第三边)解题思路
水题,啥都不说枚举即可。题目代码
#include <set> #include <map> #include <queue> #include <math.h> #include <vector> #include <string> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <iostream> #include <cctype> #include <algorithm> #include <time.h> #define eps 1e-10 #define pi acos(-1.0) #define inf 107374182 #define inf64 1152921504606846976 #define lc l,m,tr<<1 #define rc m + 1,r,tr<<1|1 #define zero(a) fabs(a)<eps #define iabs(x) ((x) > 0 ? (x) : -(x)) #define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A)))) #define clearall(A, X) memset(A, X, sizeof(A)) #define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE)) #define memcopyall(A, X) memcpy(A , X ,sizeof(X)) #define max( x, y ) ( ((x) > (y)) ? (x) : (y) ) #define min( x, y ) ( ((x) < (y)) ? (x) : (y) ) using namespace std; bool judgeTri(int a,int b,int c) { if(a+b>c&&iabs(a-b)<c)return true; return false; } bool judgeSeg(int a,int b,int c) { if(a+b==c)return true; return false; } int num[5],n=4; int main() { for(int i=0;i<n;i++) { scanf("%d",&num[i]); } bool flat1=false,flat2=false; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { for(int k=j+1;k<n;k++) { if(judgeTri(num[i],num[j],num[k])||judgeTri(num[i],num[k],num[j])||judgeTri(num[k],num[j],num[i]))flat1=true; if(judgeSeg(num[i],num[j],num[k])||judgeSeg(num[i],num[k],num[j])||judgeSeg(num[k],num[j],num[i]))flat2=true; } } } if(flat1) { puts("TRIANGLE"); } else if(flat2) { puts("SEGMENT"); } else puts("IMPOSSIBLE"); return 0; }
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