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, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

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

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

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

 int main(){
     cin >> a >> b >> c;
     int t = min(min(a / , b / ), c / );

     ans +=  * t;
     a -=  * t;
     b -=  * t;
     c -=  * t;

     rFor(i, , ) {
         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 + ;
                 i = ;
                 break;
             }

         }
     }

     cout << ans << endl;
     return ;
 }