For n edges, look for three edges, make it a triangle, and make the triangle the largest perimeter.

Analysis: Water problem, first of all the side of the sort, and then only need to determine whether the adjacent three edges can form a triangle, if possible, then the current triangle is all the circumference of the triangle before the largest, scan once sorted results can be the final answer.

#include <iostream> #include <algorithm>using namespace Std;int a[10005];int main () {int N,i,len,tmp;while ( Cin>>n && N) {len=0;for (i=0;i<n;i++) cin>>a[i];sort (a,a+n); Len=0;for (i=0;i+2<n;i++) if (a[ I]+a[i+1]>a[i+2]) {tmp=a[i]+a[i+1]+a[i+2];len=len>tmp?len:tmp;} Cout<<len<<endl;}    return 0;}

For n edges, look for three edges, make it a triangle, and make the triangle the largest perimeter.