HDU 1104 Reminder(BFS)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1104

Remainder

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1916    Accepted Submission(s): 411

Problem Description

Coco is a clever boy, who is good at mathematics. However, he is puzzled by a difficult mathematics problem. The problem is: Given three integers N, K and M, N may adds (‘+’) M, subtract (‘-‘) M, multiples (‘*’) M or modulus (‘%’) M (The definition of ‘%’ is given below), and the result will be restored in N. Continue the process above, can you make a situation that “[(the initial value of N) + 1] % K” is equal to “(the current value of N) % K”? If you can, find the minimum steps and what you should do in each step. Please help poor Coco to solve this problem.

You should know that if a = b * q + r (q > 0 and 0 <= r < q), then we have a % q = r.

 

Input

There are multiple cases. Each case contains three integers N, K and M (-1000 <= N <= 1000, 1 < K <= 1000, 0 < M <= 1000) in a single line.

The input is terminated with three 0s. This test case is not to be processed.

 

Output

For each case, if there is no solution, just print 0. Otherwise, on the first line of the output print the minimum number of steps to make “[(the initial value of N) + 1] % K” is equal to “(the final value of N) % K”. The second line print the operations to do in each step, which consist of ‘+’, ‘-‘, ‘*’ and ‘%’. If there are more than one solution, print the minimum one. (Here we define ‘+’ < ‘-‘ < ‘*’ < ‘%’. And if A = a1a2...ak and B = b1b2...bk are both solutions, we say A < B, if and only if there exists a P such that for i = 1, ..., P-1, ai = bi, and for i = P, ai < bi)

 

Sample Input

2 2 2 -1 12 10 0 0 0

 

Sample Output

0 2 *+

 

Author

Wang Yijie

 

Recommend

Eddy

 

以下解题思路来自：http://www.cnblogs.com/qiufeihai/archive/2012/08/28/2660272.html

题意：（注意题目中的%是指mod）开始给了你n, k, m。。。。每次由+m, -m, *m, modm得到新的N，继续对N这样的操作,直到(n+1) mod k== N mod k时结束。。。并且打印路径

%与mod的区别：%出来的数有正有负，符号取决于左操作数。。。而mod只能是正（因为a = b * q + r (q > 0 and 0 <= r < q), then we have a mod q = r    中r要大于等于0小于q）。。。。。

所以要用%来计算mod的话就要用这样的公式：a mod b = (a % b + b) % b

括号里的目的是把左操作数转成正数

 由于新的N可以很大，所以我们每一步都要取%，而且最后要mod k，正常来说每步都%k就行了，但是由于其中的一个操作是N%m,所以我们每一步就不能%k了（%k%m混用会导致%出来的答案错误），而要%(k *m)(其实%(k,m的公倍数都行))

然后，vis[这里放的要是遍历的点mod k (想清楚标记的目的是避免结果重复)]

而那四个操作避免过大则取余就可以了，而不需要取mod

记录路径，直接用string来累加路径就行了。。。

代码：

#include<cstdio>
#include<iostream>
#include<queue>
#include<stack>
#include<cstring>
#include<cstdlib>
using namespace std;
int n, k, m;
int vis[3002];
struct node
{
    int num;
    int time;
    string road;
};
void bfs()
{
    queue<node>q;
    memset(vis, 0, sizeof(vis));
    node p, s;
    int i;
    p.num = n;
    p.time = 0;
    p.road = "";
    vis[(n % k + k) % k] = 1;
    q.push(p);
    while(!q.empty())
    {
        p = q.front();
        q.pop();
        if((p.num % k + k) % k == ((n + 1) % k + k) % k)
        {
            printf("%d\n", p.time);
            cout << p.road << endl;
            return ;
        }
        s.time = p.time + 1;
        for(i = 0; i < 4; i++)
        {
            switch(i)
            {
            case 0:
                s.num = ((p.num + m) % (k * m) + (k * m)) % (k * m);
                s.road = p.road + '+';
                break;
            case 1:
                s.num = ((p.num - m) % (k * m) + (k * m)) % (k * m);
                s.road = p.road + '-';
                break;
            case 2:
                s.num = ((p.num * m) % (k * m) + (k * m)) % (k * m);
                s.road = p.road + '*';
                break;
            case 3:
                s.num = (p.num % m + m) % m % (k * m); //a mod b = (a % b + b) % b
                s.road = p.road + '%';
                break;
            }
            if(!vis[(s.num % k + k) % k])
            {
                vis[(s.num % k + k) % k] = 1;
                q.push(s);
            }
        }
    }
    printf("0\n");
}
int main()
{
    while(scanf("%d %d %d", &n, &k, &m) != EOF)
    {
        if(!n && !m && !k)break;
        bfs();
    }
    return 0;
}