zoukankan      html  css  js  c++  java
  • HDU 1104 Remainder

    题目链接:

    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): 3566    Accepted Submission(s): 828

    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
    *+
     
    题解:
      bfs求最短路,很直接,很暴力。但有些地方需要注意:
      正常的想法是:
        (n+m)%k=(n%k+m)%k
      这样做的话问题规模会在1000。
      但是“%”不像"+","-","*"一样,这个式子对“%”是不成立的也就是说:
        (n%m)%k != ((n%k)%m)%k=n%k%m%k
      因此为了保证n在%m操做之前不要因为%k操做而失真,我们把%k改成%km (km=k*m)
      即:
        (n%m)%k=(((n%km)%m))%km
     
    ac代码:
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<queue>
     5 #include<string>
     6 using namespace std;
     7 
     8 const int maxn = 1e3 + 10;
     9 
    10 struct Node {
    11     int val, step;
    12     string s;
    13     Node(int v, int st, string s) :val(v), step(st),s(s){}
    14     Node() {}
    15 };
    16 
    17 int n, k, m,km;
    18 
    19 int vis[maxn*maxn];
    20 
    21 void bfs(int ans) {
    22     queue<Node> q;
    23     Node node=Node(((n%km) + km) % km, 0, "");
    24     vis[node.val] = 1;
    25     q.push(node);
    26     while (!q.empty()) {
    27         node = q.front(); q.pop();
    28         if (node.val%k == ans) {
    29             cout << node.step << endl;
    30             cout << node.s << endl;
    31             return;
    32         }
    33         int tmp = (node.val + m) % km;
    34         if (!vis[tmp]) {
    35             vis[tmp] = 1;
    36             q.push(Node(tmp, node.step + 1, node.s + "+"));
    37         }
    38         tmp = ((node.val - m) % km + km) % km;
    39         if (!vis[tmp]) {
    40             vis[tmp] = 1;
    41             q.push(Node(tmp, node.step + 1, node.s + "-"));
    42         }
    43         tmp = (node.val*m) % km;
    44         if (!vis[tmp]) {
    45             vis[tmp] = 1;
    46             q.push(Node(tmp, node.step + 1, node.s + "*"));
    47         }
    48         tmp = (node.val%m) % km;
    49         if (!vis[tmp]) {
    50             vis[tmp] = 1;
    51             q.push(Node(tmp, node.step + 1, node.s + "%"));
    52         }
    53     }
    54     cout << "0" << endl;
    55 }
    56 
    57 void init() {
    58     memset(vis, 0, sizeof(vis));
    59 }
    60 
    61 int main() {
    62     while (scanf("%d%d%d", &n, &k, &m) == 3) {
    63         if (n == 0 && k == 0 && m == 0) break;
    64         km = k*m;
    65         init();
    66         bfs(((n+1)%k+k)%k);
    67     }
    68     return 0;
    69 }
    View Code
     
  • 相关阅读:
    蓝牙打印机的连接方法
    CE不能开机的可能情况
    手机性能指标的建议
    WINCE中使用键盘钩子的注意事项
    Wave接口开发注意事项
    解决唤醒屏不亮的问题之总结
    WM系统有用的注册表(研发人员使用)
    快速求解两个时间之间的天数
    测试SqlBulkCopy类批量插入数据
    Xml与DataTable相互转换方法
  • 原文地址:https://www.cnblogs.com/fenice/p/5247048.html
Copyright © 2011-2022 走看看