Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 30281		Accepted: 9124

Description

John is going on a fishing trip. He has h hours available (1 <= h <= 16), and there are n lakes in the area (2 <= n <= 25) all reachable along a single, one-way road. John starts at lake 1, but he can finish at any lake he wants. He can only travel from one lake to the next one, but he does not have to stop at any lake unless he wishes to. For each i = 1,...,n - 1, the number of 5-minute intervals it takes to travel from lake i to lake i + 1 is denoted ti (0 < ti <=192). For example, t3 = 4 means that it takes 20 minutes to travel from lake 3 to lake 4. To help plan his fishing trip, John has gathered some information about the lakes. For each lake i, the number of fish expected to be caught in the initial 5 minutes, denoted fi( fi >= 0 ), is known. Each 5 minutes of fishing decreases the number of fish expected to be caught in the next 5-minute interval by a constant rate of di (di >= 0). If the number of fish expected to be caught in an interval is less than or equal to di , there will be no more fish left in the lake in the next interval. To simplify the planning, John assumes that no one else will be fishing at the lakes to affect the number of fish he expects to catch.

Write a program to help John plan his fishing trip to maximize the number of fish expected to be caught. The number of minutes spent at each lake must be a multiple of 5.

Input

You will be given a number of cases in the input. Each case starts with a line containing n. This is followed by a line containing h. Next, there is a line of n integers specifying fi (1 <= i <=n), then a line of n integers di (1 <=i <=n), and finally, a line of n - 1 integers ti (1 <=i <=n - 1). Input is terminated by a case in which n = 0.

Output

For each test case, print the number of minutes spent at each lake, separated by commas, for the plan achieving the maximum number of fish expected to be caught (you should print the entire plan on one line even if it exceeds 80 characters). This is followed by a line containing the number of fish expected.

If multiple plans exist, choose the one that spends as long as possible at lake 1, even if no fish are expected to be caught in some intervals. If there is still a tie, choose the one that spends as long as possible at lake 2, and so on. Insert a blank line between cases.

Sample Input

2 1 10 1 2 5 2 4 4 10 15 20 17 0 3 4 3 1 2 3 4 4 10 15 50 30 0 3 4 3 1 2 3 0

Sample Output

45, 5 Number of fish expected: 31 240, 0, 0, 0 Number of fish expected: 480 115, 10, 50, 35 Number of fish expected: 724

Source

解析：由于是从第一个湖出发的。并且全部的湖都是一字排开的，所以仅仅需枚举他走过的湖泊数X就可以。即先如果他从湖1走到湖X。则路上总共花了T= T1 + T2 + T3 + ... + Tx。在这个前提下。就能够觉得他有能力在1~X之间的不论什么两个湖之间“瞬移”。即在任一时刻能够任选一个1~X中的湖钓鱼。

（想一想为什么？事实上这跟汽车加油的道理是一样的，在每一个湖的钓鱼顺序能够不是依次来的，你可能觉得总时间肯定比这个花得多。事实上不是的，顺序事实上是不影响结果的。由于假如我要先去湖1钓5分钟，接着去湖2钓5分钟。再接着回来湖1钓5分钟，这个过程事实上相当于先在湖1钓5+5=10分钟，然后再去湖2钓5分钟）。因此仅仅需一直贪心的选择当前能钓到鱼最多的湖就可以。还有就是贪心选择的时候。若有同样的湖时，优先选择编号较小的湖。

AC代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;const int maxn = 30;int t[maxn], f[maxn], d[maxn];struct node{ int id; int f; int d; friend bool operator <(node a, node b){ //注意从大到小排，要重载 '<' if(a.f == b.f) return a.id > b.id; //若鱼数相等，则选择id较小的 return a.f < b.f; }};node fish[maxn];int times[maxn][maxn]; //记录每一个湖钓鱼时间int main(){ #ifdef sxk freopen("in.txt", "r", stdin); #endif // sxk int n, h; while(scanf("%d", &n)!=EOF && n){ scanf("%d", &h); memset(times, 0, sizeof(times)); h = h * 12; for(int i=1; i<=n; i++){ scanf("%d", &fish[i].f); fish[i].id = i; } for(int i=1; i<=n; i++) scanf("%d", &fish[i].d); for(int i=1; i<=n-1; i++) scanf("%d", &t[i]); int maxans = 0; int maxk = 1; for(int i=1; i<=n; i++){ int tc = 0; for(int j=1; j
           
             p; for(int j=1; j<=i; j++) p.push(fish[j]); //将湖1~X的鱼量放入从大到小排的优先队列 int ans = 0; int t = h - tc; for(int j=1; j<=t; j++){ node foo = p.top(); ans += foo.f; times[i][foo.id] += 5; p.pop(); p.push(node{foo.id, max(foo.f - foo.d, 0), foo.d}); } if(maxans < ans){ maxans = ans; maxk = i; } } for(int i=1; i