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.
題目大意： 一隻貓只吃三種食物，在星期一、四、七喫a品種食物，星期二、六喫b品種食物，星期三、五喫c品種食物，現給出a，b，c，問最多能餵食貓幾天。
思路： 先對a，b，c的值進行判斷，顯然，當a>=3&&b>=2&&c>=2時，至少能喫到一個星期，再算剩下的最多能連續喫幾天（從星期一到星期天進行枚舉）。
#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) { // 如果給出的貓糧至少能喫一個星期，求出最多能喫幾個星期
int k = min(a / 3, min(b / 2, c / 2)); // 找出最多能喫的星期數
a -= 3 * k; // 對應的a，b，c值也要相應的改變
b -= 2 * k;
c -= 2 * k;
ans = 7 * k; // 當前能喫的天數
}
for (int i = 0; i < 7; i++) { // 從星期一到星期天進行枚舉，找出剩餘最多能連續喫幾天
g[0] = a, g[1] = b, g[2] = c;
cnt = 0;
for (int j = i; ; j++) { // 枚舉天數
if (g[date[j % 7]] > 0) g[date[j % 7]]--, cnt++; // 如果當前還能喂貓糧，該品種貓糧數目減一，天數加一
else break; // 如果當前沒有該品種貓糧，退出
}
if (cnt > maxn) maxn = cnt; // 找出最大天數
}
printf("%d", ans + maxn);
return 0;
}