zoukankan      html  css  js  c++  java
  • CCPC热身赛题解(2021广州站

    A题是2021ICPC网络预选赛第二场的M题,C题源自2013 ACM/ICPC Asia Regional Hangzhou Online
    C题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4741

    B题题目如下,代码在末尾

    Kongming’s Password

    Input file: standard input
    Output file: standard output
    Time limit: 1 second
    Memory limit: 512 megabytes

    In the ancient period of Three Kingdoms, Zhuge Kongming, the Prime Minister of Kingdom Shu, was the most famous and wise statesman. His enemy was Sima Zhongda, the Grand Preceptor of Kingdom Wei. Zhongda always looked stupid when fighting against Kongming. But it was Zhongda who laughed to the end. Kongming had led his army across the Mountain Qi to attack Kingdom Wei six times, which all failed. Because of the long journey, the food supply was a big problem.Kongming invented a kind of transportation robot called “Wooden Bull & Floating Horse” (in abbreviation,WBFH) to carry food for the army. Every WBFH had a password lock. A WBFH would move if and only if a soldier entered the correct password. Kongming was always worrying about everything and always did trivial things by himself. Since he put Ma Youchang to death due to losing the Battle of Jieting, Kongming did not trust anyone’s IQ anymore. He thought the soldiers might forget the password of WBFHs. So he made two password cards for each WBFH. If the soldier operating a WBFH forgot the password or got killed, the password still could be recovered by those two password cards.Once, Zhongda defeated Kongming again and got many WBFHs on the battlefield. But he did not know the passwords. Youchang’s son betrayed Kongming and came to Zhongda. He told Zhongda the way to figure out the password from the two cards. He said to Zhongda: “A password card is a square grid consisting of N × N cells. In each cell, there is a number. Two password cards are of the same size. If you overlap them, you get two numbers in each cell. Those two numbers in a cell may be the same or different. You can turn a card by 0 degree, 90 degrees, 180 degrees, or 270 degrees, and then overlap it on another. But flipping is not allowed. The maximum amount of cells that contain two equal numbers after overlapping, is the password. Please note that the two cards must be totally overlapped. You cannot only overlap a part of them.” Now you should find a way to figure out the password for each WBFH as quickly as possible.

    Input

    There are several test cases. In each test case:
    The first line contains an integer N (0 < N ≤ 30) — the password card is an N × N grid.
    Then an N × N matrix follows, describing one password card. Each element is an integer in a cell.
    Then another N × N matrix follows, describing the other password card.
    Those integers are all no less than 0 and less than 300.
    The input ends with N = 0.

    2
    1 2
    3 4
    5 6
    7 8
    2
    10 20
    30 13
    90 10
    13 21
    0

    Output

    For each test case, print the password.
    0
    2

    点击查看代码
    #include<bits/stdc++.h>
    using namespace std;
    
    int main() {
        int n;
        while(cin>>n,n) {
            int a[n+5][n+5],b[n+5][n+5];
            int c[n+5][n+5];
            for(int i = 0;i < n;i++) {
                for(int j = 0;j < n;j++) {
                    cin>>a[i][j];
                }
            }
            int ret = 0;
            int cnt = 0;
            for(int i = 0;i < n;i++) {
                for(int j = 0;j < n;j++) {
                    cin>>b[i][j];
                    if(a[i][j]==b[i][j]) cnt++;
                }
            }
            ret = max(cnt,ret);
            for(int k = 0;k < 3;k++) {
                for(int i = 0;i < n;i++) {
                    for(int j = 0;j < n;j++) {
                        c[i][j] = a[j][n-i-1];
                    }
                }
                for(int i = 0;i < n;i++) {
                    for(int j = 0;j < n;j++) {
                        a[i][j] = c[i][j];
                    }
                }
                /*
                for(int i = 0;i < n;i++) {
                    for(int j = 0;j < n;j++) {
                        cout<<a[i][j]<<' ';
                    }
                    cout<<endl;
                }
                */
                cnt = 0;
                for(int i = 0;i < n;i++) {
                    for(int j = 0;j < n;j++) {
                        if(a[i][j]==b[i][j]) cnt++;
                    }
                }
                ret = max(cnt,ret);
            }
        }
        return 0;
    }
    /*
    逆时针旋转
    1 2
    3 4
    
    3 1
    4 2
    
    4 3
    2 1
    
    2 4
    1 3
    -----
    1 2
    3 4
    */
    
    
  • 相关阅读:
    8.9乘船问题
    8.8几个背包问题
    8.7贪心策略例题:字典序最小问题
    8.6贪心策略例题:区间覆盖问题
    8.5贪心策略例题:区间选点问题
    8.4贪心策略例题:区间调度问题
    SQL 报表 生成月份临时表
    CentOS8 .NET Core项目部署
    Django with database on SQL Server
    SQL 父子表,显示表中每条记录所在层级
  • 原文地址:https://www.cnblogs.com/R1ST/p/15549025.html
Copyright © 2011-2022 走看看