hdu 1104 数论+bfs

Remainder

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2122 Accepted Submission(s): 449

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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)

[align=left]Sample Input[/align]

2 2 2
-1 12 10
0 0 0

[align=left]Sample Output[/align]

0
2
*+

/*
其他没什么好说的，数字太大需要%（k*m）,这个可以证明就等于对n进行+、-m操作
不影响结果，属于正常操作
*/
#include<iostream>
#include<queue>
#include<string>
using namespace std;
bool vis[1000010];

struct point
{
int val;
int step;
string s;
}p,t;

void bfs(int n,int k,int m)
{
memset(vis,false,sizeof(vis));
queue<point> q;
int s=((n+1)%k+k)%k;
t.val=n;
t.step=0;
t.s="";
vis[(n%k+k)%k]=true;
q.push(t);
while(!q.empty())
{
t=q.front();
q.pop();
if(s==(t.val%k+k)%k)
{
cout<<t.step<<endl;
cout<<t.s<<endl;
return ;
}
for(int i=0;i<4;i++)
{
p=t;
p.step++;
if(i==0)
{
p.val=(t.val+m)%(k*m);
p.s+='+';
}
else if(i==1)
{
p.val=(t.val-m)%(k*m);
p.s+='-';
}
else if(i==2)
{
p.val=(t.val*m)%(k*m);
p.s+='*';
}
else if(i==3)
{
p.val=(t.val%m+m)%m%(k*m);
p.s+='%';
}
if(!vis[(p.val%k+k)%k])
{
q.push(p);
vis[(p.val%k+k)%k]=true;
}
}
}
cout<<0<<endl;
}
int main()
{
int n,m,k;
while(cin>>n>>k>>m,k || m || n)
bfs(n,k,m);
return 0;
}