Codeforces Div.3 C. Gourmet Cat——————思维，贪心

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本文介绍了一种算法，帮助猫主人Polycarp根据背包中鱼食、兔肉汤和鸡腿的每日配给数量，选择最佳出发日，使他的猫GourmetCat在旅行期间无需额外购买食物的情况下尽可能长时间地享受美食。

C. Gourmet Cat

Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:

on Mondays, Thursdays and Sundays he eats fish food;
on Tuesdays and Saturdays he eats rabbit stew;
on other days of week he eats chicken stake.

Polycarp plans to go on a trip and already packed his backpack. His backpack contains:

a daily rations of fish food;
b daily rations of rabbit stew;
c daily rations of chicken stakes.

Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.

Input
The first line of the input contains three positive integers a, b and c $1≤a,b,c≤7⋅10^8)$ — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.

Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.

Examples
input
2 1 1
output
4
input
3 2 2
output
7
input
1 100 1
output
3
input
30 20 10
output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.

In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.

In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.

简单思维
可以发现，
a:1,4,7
b:2,6
c:3,5
先把完整的星期去掉，然后再枚举开始时间就好了

#include<bits/stdc++.h>
// #define SUBMIT
using namespace std;
typedef long long ll;
int main()
{
    #ifdef SUBMIT
    freopen("../in.txt", "r", stdin);
    freopen("../out.txt", "w", stdout);
    ll _begin_time = clock();
    #endif

    int a,b,c;
    while(cin>>a>>b>>c)
    {
           // cin>>a>>b>>c;
            int ans = 0;

            int minn = min(a/3,min(b/2,c/2));
            ans = minn*7;
            a -= minn*3;
            b -= minn*2;
            c -= minn*2;
            int p = 0;
            for(int i=1;i<=7;++i)
            {
                    int t = i;//枚举开始时间
                    int maxx = 0;//以星期i 开始，能吃的最多天数
                    int aa = a, bb=b, cc=c;
                    while(1)
                    {
                            if(t%7==1||t%7==4||t%7==0)  aa--;
                            if(t%7==2||t%7==6)          bb--;
                            if(t%7==3||t%7==5)          cc--;
                            t++;
                            maxx++;
                            if(aa<0||bb<0||cc<0)//如果下一天还能吃，那就继续循环，否则结束循环
                            {
                                    if(p<maxx)  p = maxx;
                                    break;
                            }
                    }
            }
            cout<<ans+p-1<<endl;
    }

    #ifdef SUBMIT
    ll _end_time = clock();
    printf("\ntime = %I64d ms", _end_time - _begin_time);
    #endif
    return 0;
}