猫咪饮食计划算法解析

https://codeforces.com/contest/1154/problem/C

题意：你有只猫，周一、四、日吃鱼，周二、六吃兔肉。其余时间吃鸡肉。现在你要去旅行，且你准备了a份鱼、b份兔肉、c份鸡肉。求你在某天出门后这只猫能吃的的最大天数。

题解：枚举

1、枚举开始的星期几；

2、规律（这个天数跟你周几出门有关

#include <bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<string>

#include<vector>

#include<bitset>

#include<queue>

#include<deque>

#include<stack>

#include<cmath>

#include<list>

#include<map>

#include<set>

#define RI register int

#define endl "n"

using namespace std;

typedef long long ll;

const int N=100000+10;

const int M=100000+10;

const int MOD=1e9+7;

const double PI = acos(-1.0);

const double EXP = 1E-8;

const int INF = 0x3f3f3f3f;

int t,n,m,k,p,l,r,u,v;

ll ans,cnt,flag,temp,sum;

int a[4],b[8]={0,1,2,3,1,3,2,1},c[4];

char str;

struct node{};

ll checked(int week){

ll day=0;

for(int i=1;i<=3;i++)

c[i]=a[i];

for(int i=week;i<=7;i++){

if(c[b[i]])c[b[i]]--,day++;

else return i-week;

}

ll mid=min(c[1]/3,min(c[2]/2,c[3]/2));

day+=(mid*7);

c[1]-=(mid*3);

c[2]-=(mid*2);

c[3]-=(mid*2);

for(int i=1;i<=7;i++){

if(c[b[i]])c[b[i]]--,day++;

else break;

}

return day;

}

int main()

{

#ifdef DEBUG

freopen("input.in", "r", stdin);

#endif

scanf("%d%d%d",&a[1],&a[2],&a[3]);

for(int i=1;i<=7;i++){

ans=max(ans,checked(i));

}

cout<<ans<<endl;

#ifdef DEBUG

printf("Time cost : %lf sn",(double)clock()/CLOCKS_PER_SEC);

#endif

return 0;

}

cpp

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