hdu1104Remainder （BFS）
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.
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.
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 *+
看了他人的才AC，自己写的会超内存！！下面 的话是从他人的博客里面复制过来的，看起来还不错，就是有一点不明，为什么要余K*M（或是它们的公陪数）？
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。。。
#include<iostream>
#include<queue>
#include<stdio.h>
#include<cstring>
using namespace std;
typedef struct nn
{
string way;
int step,n;
}node;
int N,K,M,vist[1000000];
void BFS()
{
queue<node>Q;
node q,p;
memset(vist,0,sizeof(vist));
int i,km=K*M,mod=((N+1)%K+K)%K;//（N+1）%K余数要为正，为负时变正
vist[(N%K+K)%K]=1; //记录余数有没有访问过
for(i=0;i<4;i++)//一定要按照顺序更新n，因为在题目给定时，要求路径最小
{
q.way="";
if(i==0)
{ q.n=(N+M)%km;q.way+='+';q.step=1;}
if(i==1)
{ q.n=(N-M)%km;q.way+='-';q.step=1;}
if(i==2)
{ q.n=(N*M)%km;q.way+='*';q.step=1;}
if(i==3)
{ q.n=((N%M+M)%M)%km;q.way+='%';q.step=1;}
if(vist[(q.n%K+K)%K]==0)
Q.push(q);
vist[(q.n%K+K)%K]=1;
}
while(!Q.empty())
{
q=Q.front();Q.pop();
if((q.n%K+K)%K==mod)//找到
{
printf("%d\n",q.step);
for(i=0;i<q.step;i++)
printf("%c",q.way[i]);
printf("\n");
return ;
}
for(i=0;i<4;i++)
{
p.way=q.way;
if(i==0)
{
p.n=(q.n+M)%km;//最km的余是因为如果不最余，那么n的值会很大
p.way+='+';p.step=1+q.step;
}
if(i==1)
{
p.n=(q.n-M)%km;p.way+='-';p.step=1+q.step;
}
if(i==2)
{
p.n=(q.n*M)%km;p.way+='*';p.step=1+q.step;
}
if(i==3)
{
p.n=((q.n%M+M)%M)%km;p.way+='%';p.step=1+q.step;
}
if(vist[(p.n%K+K)%K]==0)//当不存在时可以放入，存在时如果放入那下次取出会重复和余数相同的步聚
Q.push(p);
vist[(p.n%K+K)%K]=1;
}
}
printf("0\n");
}
int main()
{
int i;
while(scanf("%d%d%d",&N,&K,&M)>0&&(N||M||K))
{
BFS();
}
}