Codeforces 1154C Gourmet Cat

题目链接：http://codeforces.com/problemset/problem/1154/C

题目大意：

主人有一只猫。
周一&周四&周日：吃鱼
周二&周六：吃兔子
周三&周五：吃鸡

他们现在要外出旅游。他们带了一个包，包里有：
a份鱼肉
b份兔肉
c份鸡肉

问，猫的主人在最优解的情况下（即他可以自由的选择周几出去旅游），最多可以多少天不额外在外购买猫粮？（即，这个包最多能够猫吃多少天）

【猫一天就吃一份】

分析：

　　首先把能吃完整7天的食量减掉，接下来的食物配比就只能吃0~6天了，可以把满足能吃i（0 < i < 7）天的所有实物配比情况全部列出来，再枚举，能大大减少代码量。

代码如下：

#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl

#define LOWBIT(x) ((x)&(-x))

#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())

#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define pii pair<int,int> 
#define piii pair<pair<int,int>,int> 
#define mp make_pair
#define pb push_back
#define fi first
#define se second

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

typedef long long LL;
typedef unsigned long long uLL;
const double EPS = 1e-9;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;

struct Food{
    int x, y, z;
};

int n, a, b, c, ans; 
vector< Food > foods[6] = {
                            {Food(0, 0, 1), Food(0, 1, 0), Food(1, 0, 0)}, 
                            {Food(2, 0, 0), Food(0, 1, 1), Food(1, 1, 0), Food(1, 0, 1)}, 
                            {Food(2, 1, 0), Food(1, 1, 1), Food(1, 0, 2)}, 
                            {Food(2, 2, 0), Food(2, 1, 1), Food(1, 1, 2)}, 
                            {Food(2, 2, 1), Food(2, 1, 2), Food(1, 2, 2)}, 
                            {Food(2, 2, 2), Food(3, 1, 2), Food(3, 2, 1)}
                        };

int main(){
    cin >> a >> b >> c;
    int t = min(min(a / 3, b / 2), c / 2);
    
    ans += 7 * t;
    a -= 3 * t;
    b -= 2 * t;
    c -= 2 * t;
    
    rFor(i, 5, 0) {
        foreach(j, foods[i]) {
            if(a >= j->x && b >= j->y && c >= j->z) {
                a -= j->x; b -= j->y; c -= j->z;
                ans += i + 1;
                i = 0;
                break;
            }
            
        }
    }
    
    cout << ans << endl;
    return 0;
}