Fzu 2127 Chicken Farm

Topic Link: Chicken Farm

This is probably the two days to do the most smooth together, after reading the 1 A. However, the reading of the puzzle.

Feeling is a maths problem.

Start thinking about if the two-edge enumeration range is definitely timed out. But like this, paper-cut is still hopeful.

Enumerates the extents of the first edge, finds the number of possible values for the second edge on each value of the first edge, and then finds the number of possible values for the third edge B.

In this case, the number of triangles that each value can make is a small value between A and B. "For what?" it's obvious.

Code knowledge is only used on both sides of the sum greater than the third side, this range of limits of thought worth learning ...

Attached code:

#include <stdio.h> #include <string.h> #include <iostream>using namespace Std;int main () {    int n;< C1/>int L1, R1, L2, R2, L3, R3;    while (CIN >> N) {        int ans = 0;        CIN >> L1 >> R1 >> L2 >> R2 >> L3 >> R3;        for (int i=l1; i<=r1; ++i) {            int temp = (n-i)/2;            if (i > Temp) break;            A2 left Border            int min2 = max (L2, I);//a2>a1            min2 = max (min2, (n-i*2)/2+1);//A3<A1+A2            //A2 right boundary            int max2 = min (r2, (n-i)/2); A1<A3            //A3 left Border            int min3 = max (L3, N-I-MAX2);//a3>a2            int max3 = min (r3, n-i-min2);//a3>a2
   
    temp = min (max2-min2+1, max3-min3+1);            if (temp>0)                ans + = temp;        }        printf ("%d\n", ans);    }    return 0;}
   

　　

Fzu 2127 Chicken Farm