「CF452D」Washer, Dryer, Folder
You have k pieces of laundry, each of which you want to wash, dry and fold. You are at a laundromat that has n1 washing machines,n2 drying machines and n3 folding machines. Each machine can process only one piece of laundry at a time. You can’t dry a piece of laundry before it is washed, and you can’t fold it before it is dried. Moreover, after a piece of laundry is washed, it needs to be immediately moved into a drying machine, and after it is dried, it needs to be immediately moved into a folding machine.
It takes t1 minutes to wash one piece of laundry in a washing machine, t2 minutes to dry it in a drying machine, and t3 minutes to fold it in a folding machine. Find the smallest number of minutes that is enough to wash, dry and fold all the laundry you have.
The only line of the input contains seven integers: k, n1, n2, n3, t1, t2, t3 (1 ≤ k ≤ 104; 1 ≤ n1, n2, n3, t1, t2, t3 ≤ 1000).
Print one integer — smallest number of minutes to do all your laundry.
|
1 |
1 1 1 1 5 5 5 |
|
1 |
15 |
|
1 |
8 4 3 2 10 5 2 |
|
1 |
32 |
In the first example there’s one instance of each machine, each taking 5 minutes to complete. You have only one piece of laundry, so it takes 15 minutes to process it.
In the second example you start washing first two pieces at moment 0. If you start the third piece of laundry immediately, then by the time it is dried, there will be no folding machine available, so you have to wait, and start washing third piece at moment 2. Similarly, you can’t start washing next piece until moment 5, since otherwise there will be no dryer available, when it is washed. Start time for each of the eight pieces of laundry is 0, 0, 2, 5, 10, 10, 12 and 15 minutes respectively. The last piece of laundry will be ready after 15 + 10 + 5 + 2 = 32 minutes.
题解
k件衣服，可以同时洗n0件，同时晒n1,同时叠n2，洗一件的时间t0,晒t1,叠t2，一件衣服洗完要马上晒，晒完马上叠，问处理k件衣服的最小时间
递推
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 |
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #define ll long long using namespace std; inline ll read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int t[10005],k,n0,n1,n2,t0,t1,t2; int main() { k=read();n0=read();n1=read();n2=read();t0=read();t1=read();t2=read(); for(int i=0;i<k;i++) { if(i>=n0)t[i]=max(t[i],t[i-n0]+t0); if(i>=n1)t[i]=max(t[i],t[i-n1]+t1); if(i>=n2)t[i]=max(t[i],t[i-n2]+t2); } printf("%d\n",t[k-1]+t0+t1+t2); return 0; } |
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