Mahmoud has n line segments, the i-th of them has length ai. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn’t accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can’t concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
InputThe first line contains single integer n (3?≤?n?≤?105) — the number of line segments Mahmoud has.
The second line contains n integers a1,?a2,?…,?an (1?≤?ai?≤?109) — the lengths of line segments Mahmoud has.
OutputIn the only line PRint “YES” if he can choose exactly three line segments and form a non-degenerate triangle with them, and “NO” otherwise.
ExamplesInput51 5 3 2 4OutputYESInput34 1 2OutputNONoteFor the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.
題意: 給一組數,從中挑選三個數字是否為三角形(只要有一組符合就可以 解析: 利用兩邊之和第三邊就可以
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 100005int arr[N]; bool cmp(int x,int y){ return x > y;} int main(){ int n; while(~scanf("%d",&n)) { bool flag = false; for(int i = 0;i < n; i++) { scanf("%d",&arr[i]); } sort(arr,arr+n,cmp); for(int i = 0;i < n-2; i++) { if(arr[i] < arr[i+1]+arr[i+2]) { printf("YES/n"); flag = true; break; } } if(!flag) { printf("NO/n"); } }return 0;}
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