codeforces A. Sweet Problem

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本文解析了CodeForces竞赛中的一道A级题目，详细介绍了如何通过观察规律找到解决问题的方法，提供了完整的代码实现，适合初学者理解并掌握算法题解的思路。

原题链接：http://codeforces.com/contest/1263/problem/A

题目要求

You have three piles of candies: red, green and blue candies:

the first pile contains only red candies and there are rr candies in it,
the second pile contains only green candies and there are gg candies in it,
the third pile contains only blue candies and there are bb candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can’t eat two candies of the same color in a day.

Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.

Input
The first line contains integer tt (1≤t≤10001≤t≤1000) — the number of test cases in the input. Then tt test cases follow.

Each test case is given as a separate line of the input. It contains three integers rr, gg and bb (1≤r,g,b≤1081≤r,g,b≤108) — the number of red, green and blue candies, respectively.

Output
Print tt integers: the ii-th printed integer is the answer on the ii-th test case in the input.

Example
input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
output
1
2
2
10
5
9

解题思路

题目很简单，只要你多找几个规律，例如：
1 1 1 ： 1+1=2
1 2 1（1 1 2）：1+1=2
4 1 1（1 1 4）：1+1=2
7 4 10（4 7 10）：（4+7+10）/2=10
8 1 4（1 4 8）：1+4=5
8 2 8（2 8 8）：（2+8+8）/2=9
看出规律来了吗？那就是：只要较小的两堆糖之和大于最大的那一堆糖，我们总是可以把他吃到“0 0 0”或者“0 0 1”这种状态。所以，我们要对这种情况进行“(a[0]+a[1]+a[2])/2”这一操作。
对其他情况进行“a[0]+a[1]”操作。

正确代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string>
#include<cstring>


using namespace std;
typedef long long ll;
const ll maxn=1e8+7;
int a[5];

int main()
{
    int  t;
    scanf("%d",&t);
    while(t--){
        memset(a,0,sizeof(a));
        for(int i=0;i<3;i++){
            scanf("%d",&a[i]);
        }
        sort(a,a+3);
        if(a[0]+a[1]<=a[2])
            printf("%d\n",a[0]+a[1]);
        else
            printf("%d\n",(a[1]+a[0]+a[2])/2);
    }
    return 0;
}