湖南省第九届大学生计算机程序设计竞赛 Interesting Calculator

Interesting Calculator

Time Limit: 2 Sec  Memory Limit: 128 MB

Submit: 163  Solved: 49

Description

There is an interesting calculator. It has 3 rows of buttons.





 





Row 1: button 0, 1, 2, 3, ..., 9. Pressing each button appends that digit to the end of the display.





Row 2: button +0, +1, +2, +3, ..., +9. Pressing each button adds that digit to the display.





Row 3: button *0, *1, *2, *3, ..., *9. Pressing each button multiplies that digit to the display.





 





Note that it never displays leading zeros, so if the current display is 0, pressing 5 makes it 5 instead of 05. If the current display is 12, you can press button 3, +5, *2 to get 256. Similarly, to change the display from 0 to
1, you can press 1 or +1 (but not both!).





 





Each button has a positive cost, your task is to change the display from x to y with minimum cost. If there are multiple ways to do so, the number of presses should be minimized.

Input

There will be at most 30 test cases. The first line of each test case contains two integers x and y(0<=x<=y<=105). Each of the 3 lines contains 10 positive integers (not greater than 105), i.e. the costs of each button.





Output

For each test case, print the minimal cost and the number of presses.

Sample Input

12 256

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

12 256

100 100 100 1 100 100 100 100 100 100

100 100 100 100 100 1 100 100 100 100

100 100 10 100 100 100 100 100 100 100

Sample Output

Case 1: 2 2

Case 2: 12 3

AC代码例如以下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define M 10010
using namespace std;

struct H
{
    friend bool operator <(const H x,const H y)
    {
        return x.a>y.a||(x.a==y.a&&x.b>y.b);
    }
    int a,b,c;

};

int cas=0;
int n,m;
int r1[10],r2[10],r3[10];
int vis[100005];

void bfs()
{
    int i,j;
    H aa,bb,cc;
    priority_queue<H> q;
    aa.c=n;
    aa.a=0;
    aa.b=0;
    q.push(aa);
    while(!q.empty())
    {
        bb=q.top();
        q.pop();
        if(vis[bb.c])
            continue;
        vis[bb.c]=1;
        if(bb.c==m)
        {
            printf("Case %d: %d %d\n",cas,bb.a,bb.b);
            return;
        }
        for(i=0;i<10;i++)
        {
            cc.c=bb.c*10+i;
            if(cc.c>=0&&cc.c<100001&&!vis[cc.c])
            {
                cc.b=bb.b+1;
                cc.a=bb.a+r1[i];
                q.push(cc);
            }
            cc.c=bb.c+i;
            if(cc.c>=0&&cc.c<100001&&!vis[cc.c])
            {
                cc.b=bb.b+1;
                cc.a=bb.a+r2[i];
                q.push(cc);
            }
            cc.c=bb.c*i;
            if(cc.c>=0&&cc.c<100001&&!vis[cc.c])
            {
                cc.b=bb.b+1;
                cc.a=bb.a+r3[i];
                q.push(cc);
            }
        }
    }
}

int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        cas++;
        for(i=0;i<10;i++)
            scanf("%d",&r1[i]);
        for(i=0;i<10;i++)
            scanf("%d",&r2[i]);
        for(i=0;i<10;i++)
            scanf("%d",&r3[i]);
        memset(vis,0,sizeof vis);

        bfs();
    }
    return 0;
}