codeforces round #552(div 3) C.Gourmet Cat

C. Gourmet Cattime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarp 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.InputThe 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.OutputPrint 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.ExamplesinputCopy
2 1 1
outputCopy
4
inputCopy
3 2 2
outputCopy
7
inputCopy
1 100 1
outputCopy
3
inputCopy
30 20 10
outputCopy
39
NoteIn 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.

题解：直接暴力肯定超时，首先要先把发把他们按星期约去，在把一星期的每一天开头都测试一遍取能跑到的最大值。那么这里使用小贪心加暴力，计算过时间是够的。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main() {
 int a, b, c;
 int ans = 0;
 scanf("%d%d%d", &a, &b, &c);
 while (true) {
  if (a <= 3 || b <= 2 || c <= 2) break;
  a -= 3;
  b -= 2;
  c -= 2;
  ans += 7;
 }
 int p = 0;
 for (int i = 1; i <= 7; i++) {
  int t = i;
  int max = 0;
  int aa = a, bb = b, cc = c;
  while (true) {
   if (aa < 0 || bb < 0 || cc < 0) {
    if (max > p) {
     p = max;
    }
    break;
   }
   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++;
   max++;
  }
 }
 printf("%d\n", p + ans-1);
 return 0;
}