Codeforces Round #552 (Div. 3) C题

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

题目

C. Gourmet Cat

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

aa daily rations of fish food;
bb daily rations of rabbit stew;
cc 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 aa, bb and cc (1≤a,b,c≤7⋅1081≤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

Copy

2 1 1

output

Copy

input

Copy

3 2 2

output

Copy

input

Copy

1 100 1

output

Copy

input

Copy

30 20 10

output

Copy

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 9999 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.

思路一星期a有3次，b有2次，c有2次，所以依次相除，在选取最小表示最多坚持完整的星期有几个，之后枚举判断一星期之内(不包含7)的最多坚持天数相加即可。

AC代码

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
using namespace std;
int s[7]={1,2,3,1,3,2,1};
int main()
{int a,b,c;while(cin>>a>>b>>c){int maxn=min(min(a/3,b/2),c/2);int ans;ans=maxn*7;int A=-1;a-=3*maxn;b-=2*maxn;c-=2*maxn;for(int i=0;i<7;i++){int a1=a;int b1=b;int c1=c;int flog=0;for(int j=0;j<7;j++){if(s[(i+j)%7]==1){if(a1==0)flog=1;a1--;}else if(s[(i+j)%7]==2){if(b1==0)flog=1;b1--;}else{if(c1==0)flog=1;c1--;}if(flog==1){A=max(A,j);break;}}}ans=ans+A;cout<<ans<<endl;}return 0;
}

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