HDUOJ 1104 Remainder（广搜）

传送门
题意：等到n，k, m，每次对n用m加减乘取模操作，问至少有多少步可以使
（n的初值+1）%k与当前n%k值相等，并输出每次的运算步骤
solution：典型的广搜题目诶

#include <bits/stdc++.h>
using namespace std;

int n, k, m, km;
bool book[1000001];//标记每个点是否被走过
char C[5] = {"+-*%"};

struct abc{
	int x, step;
	string s;
};

//取模不等于%，%允许负数，取模必须是正数
int mod(int x, int y)
{
	return (x % y + y) % y;
}

int get(int x, int y)
{
	if (y == 0)return mod(x + m, km);
	if (y == 1)return mod(x - m, km);
	if (y == 2)return mod(x * m, km);
	if (y == 3)return mod(x % m, km);
}

void bfs(int z)
{
	fill(book, book + 1000001, false);
	queue<abc> q;
	km = k * m;
	n = mod(n, km);
	book[n] = true;
	q.push(abc{n, 0, ""});
	while (!q.empty()){
		abc temp = q.front();
		q.pop();
		if (mod(temp.x, k) == z){
			cout << temp.step << endl << temp.s << endl;
			return;
		}
		for (int i = 0; i < 4; ++i){
			int x = get(temp.x, i);
			if (!book[x]){
				book[x] = true;
			}else continue;
			q.push(abc{x, temp.step + 1, temp.s + C[i]});
		}
	}
	cout << 0 << endl;
}

int main()
{
	while (scanf("%d%d%d", &n, &k, &m), n + k + m)bfs(mod(n + 1, k));
	return 0;
}