A Famous ICPC Team

Time Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Mr. B, Mr. G, Mr. M and their coach Professor S are planning their way to Warsaw for the ACM-ICPC World Finals. Each of the four has a square-shaped suitcase with side length A_i (1 <= i <= 4) respectively. They want to pack their suitcases into a large square box. The heights of the large box as well as the four suitcases are exactly the same. So they only need to consider the large box’s side length. Of course, you should write a program to output the minimum side length of the large box so that the four suitcases can be put into the box without overlapping.

Input

Each test case contains only one line containing 4 integers A_i (1<= i <=4, 1<= A_i <=1,000,000,000) indicating the side length of each suitcase.

Output

For each test case, display a single line containing the case number and the minimum side length of the large box required.

Example

Input: 2 2 2 2
2 2 2 1 
Output: Case 1: 4
Case 2: 4

Explanation

For the first case, all suitcases have size 2x2. So they can perfectly be packed in a 4x4 large box without wasting any space.

For the second case, three suitcases have size 2x2 and the last one is 1x1. No matter how to rotate or move, you could find the side length of the box must be at least 4.

思路: 其实就是求最长的两个边长之和，

给出四个正方体箱子的边长，问能装下这四个正方体箱子的大正方体边长最小要多大

要边长最小且必须能装下四个箱子，那么这四个箱子肯定要叠放两层，所以边长由最长的两个小正方体边长决定

代码：

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int x[4];
    int t=1;
    while(cin>>x[0]>>x[1]>>x[2]>>x[3])
    {
        sort(x,x+4);
        cout<<"Case "<<t++<<": "<<x[2]+x[3]<<endl;
    }
    return 0;
}

#include<stdio.h>
#include<stdlib.h>
int cmp(const void *a,const void *b)
{
    return *(int *)a-*(int *)b;
}
int main(){
    int side[4];
    int i=1;
    while(~scanf("%d",&side[0]))
    {
        for(int j=1;j<4;j++)
            {
                scanf("%d",&side[j]);
            }
            qsort(side,4,sizeof(side[0]),cmp);
            printf("Case %d: ",i);
            printf("%d
",side[2]+side[3]);
            i++;
    }
return 0;
}