题目：​​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): 3320    Accepted Submission(s): 752


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 *+

分析：%与mod的区别：%出来的数有正有负，mod只能是正，即(a+mod)%mod，本题相关运算是指后者。关键词：最少的步骤，输出运算过程中的符号，相同运算次数的情况下符号序列有优先级之分。四项运算的过程中为了不产生溢出数所以进行了%km操作，还要注意每一步的%m，+m，%k，+k等操作。（WA了半天||-_-）


#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int md=1005;
int n,k,m,km;
bool vis[1000010];
struct node{
    int val,step,top;
    char path[1000];
    void init(){
        val=step=top=0;
        memset(path,0,sizeof(path));
    }
};
node que[md];
void bfs(){
    int q1=0,q2=0;
    node t;
    t.init();
    t.val=n;
    que[q2++]=t;
    vis[(t.val%k+k)%k]=1;
    while(q1!=q2){
        node cur=que[q1],temp;
        q1=(q1+1)%md;
        if(((n+1)%k+k)%k==(cur.val%k+k)%k){
            printf("%d\n%s\n",cur.step,cur.path);
            return ;
        }
        for(int i=0;i<4;i++){
            temp.init();
            strcpy(temp.path,cur.path);
            temp.top=cur.top;
            if(i==0){
                temp.val=(cur.val+m)%km;
                temp.path[temp.top++]='+';
            }
            else if(i==1){
                temp.val=(cur.val-m+km)%km;
                temp.path[temp.top++]='-';
            }
            else if(i==2){
                temp.val=cur.val*m%km;
                temp.path[temp.top++]='*';
            }
            else {
                temp.val=(cur.val%m+m)%m%km;
                temp.path[temp.top++]='%';
            }
            temp.step=cur.step+1;
            if(!vis[(temp.val%k+k)%k]){
                que[q2]=temp;
                q2=(q2+1)%md;
                vis[(temp.val%k+k)%k]=1;
            }
        }
    }
    printf("0\n");  //队列空了结束循环就是没有找到
}
int main()
{
    //freopen("cin.txt","r",stdin);
    while(cin>>n>>k>>m){
        if(n==0&&m==0&&k==0) break;
        km=k*m;
        memset(vis,0,sizeof(vis));
        bfs();
    }
    return 0;
}