hdu Remainder

Remainder

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 240    Accepted Submission(s): 65

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

分析：bfs类题，关键在于对k*m取模来作为标记下标，为避免模为负数，应使用x%y=(x%y+y)%y这个技巧。

#include
#include
#include
#include
using namespace std;
#define MAXN 1000010
int c[MAXN];
int fa[MAXN];
int vis[MAXN];
int s[MAXN];
char op[5] = "+-*%";

int mod(int a, int b) {
    return ( a % b + b) % b;
}

int main() {
    int des, n, k, m, next, now, mo, cnt;
    while (scanf("%d%d%d", &n, &k, &m) != EOF && k) {
        mo = m*k;
        queue <int> q;
        des = mod(n + 1, k);
        n = mod(n, mo);
        memset(vis, 0, sizeof (vis));
        q.push(n);
        vis[n] = 1;
        fa[n] = -1;
        c[n] = -1;
        while (!q.empty()) {
            now = q.front();
            q.pop();
            for (int i = 0; i < 4; ++i) {
                if (i == 0)next = (now + m) % mo;
                else if (i == 1)next = mod(now - m, mo);
                else if (i == 2)next = now * m % mo;
                else next = now % m;
                if (!vis[next]) {
                    vis[next] = 1;
                    fa[next] = now;
                    c[next] = i;
                    if (next % k == des)
                        goto L;
                    q.push(next);
                }
            }
        }
        printf("0\n");
        continue;
L:
        cnt = 0;
        while (fa[next] >= 0) {
            s[cnt++] = c[next];
            next = fa[next];
        }
        printf("%d\n", cnt);
        while (cnt)
            printf("%c", op[s[--cnt]]);
        printf("\n");

    }
    return 0;
}

 

转载于:https://www.cnblogs.com/baidongtan/archive/2012/09/01/2666255.html