1154C C. Gourmet Cat（枚举）

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⋅108) — 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.
题目大意： 一只猫只吃三种食物，在星期一、四、七吃a品种食物，星期二、六吃b品种食物，星期三、五吃c品种食物，现给出a，b，c，问最多能喂食猫几天。
思路： 先对a，b，c的值进行判断，显然，当a>=3&&b>=2&&c>=2时，至少能吃到一个星期，再算剩下的最多能连续吃几天（从星期一到星期天进行枚举）。

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int a, b, c, ans = 0, cnt, date[7] = {0, 1, 2, 0, 2, 1, 0}, g[3], maxn = -1;
	scanf("%d %d %d", &a, &b, &c);
	if (a >= 3 && b >= 2 && c >= 2) { // 如果给出的猫粮至少能吃一个星期，求出最多能吃几个星期 
		int k = min(a / 3, min(b / 2, c / 2)); // 找出最多能吃的星期数 
		a -= 3 * k; // 对应的a，b，c值也要相应的改变 
		b -= 2 * k;
		c -= 2 * k;
		ans = 7 * k; // 当前能吃的天数 
	}
	for (int i = 0; i < 7; i++) { // 从星期一到星期天进行枚举，找出剩余最多能连续吃几天 
		g[0] = a, g[1] = b, g[2] = c;
		cnt = 0;
		for (int j = i; ; j++) { // 枚举天数 
			if (g[date[j % 7]] > 0) g[date[j % 7]]--, cnt++; // 如果当前还能喂猫粮，该品种猫粮数目减一，天数加一 
			else break; // 如果当前没有该品种猫粮，退出 
		}
		if (cnt > maxn) maxn = cnt; // 找出最大天数 
	}
	printf("%d", ans + maxn);
	return 0;
}