CodeForce 898 C. Phone Numbers

本文介绍了一个整理电话簿的算法题目,任务是合并重复的名字条目并去除尾号重复的电话号码,最终输出整理后的电话簿。

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C. Phone Numbers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Vasya has several phone books, in which he recorded the telephone numbers of his friends. Each of his friends can have one or several phone numbers.

Vasya decided to organize information about the phone numbers of friends. You will be given n strings — all entries from Vasya's phone books. Each entry starts with a friend's name. Then follows the number of phone numbers in the current entry, and then the phone numbers themselves. It is possible that several identical phones are recorded in the same record.

Vasya also believes that if the phone number a is a suffix of the phone number b (that is, the number b ends up with a), and both numbers are written by Vasya as the phone numbers of the same person, then a is recorded without the city code and it should not be taken into account.

The task is to print organized information about the phone numbers of Vasya's friends. It is possible that two different people have the same number. If one person has two numbers x and y, and x is a suffix of y (that is, y ends in x), then you shouldn't print number x. If the number of a friend in the Vasya's phone books is recorded several times in the same format, it is necessary to take it into account exactly once.

Read the examples to understand statement and format of the output better.

Input

First line contains the integer n (1 ≤ n ≤ 20) — number of entries in Vasya's phone books.

The following n lines are followed by descriptions of the records in the format described in statement. Names of Vasya's friends are non-empty strings whose length does not exceed 10. They consists only of lowercase English letters. Number of phone numbers in one entry is not less than 1 is not more than 10. The telephone numbers consist of digits only. If you represent a phone number as a string, then its length will be in range from 1 to 10. Phone numbers can contain leading zeros.

Output

Print out the ordered information about the phone numbers of Vasya's friends. First output m — number of friends that are found in Vasya's phone books.

The following m lines must contain entries in the following format "name number_of_phone_numbers phone_numbers". Phone numbers should be separated by a space. Each record must contain all the phone numbers of current friend.

Entries can be displayed in arbitrary order, phone numbers for one record can also be printed in arbitrary order.

Examples
input
2
ivan 1 00123
masha 1 00123
output
2
masha 1 00123 
ivan 1 00123 
input
3
karl 2 612 12
petr 1 12
katya 1 612
output
3
katya 1 612 
petr 1 12 
karl 1 612 
input
4
ivan 3 123 123 456
ivan 2 456 456
ivan 8 789 3 23 6 56 9 89 2
dasha 2 23 789
output
2
dasha 2 23 789 
ivan 4 789 123 2 456 

题意:整理一个电话簿,要求:同名字的电话归在一起,同一个人的电话号码中如果一个电话是另一个电话的尾号,这个电话就舍去,相同的电话号码保留一个。最后输出整理后的电话簿。

纯模拟题,先将同名字的电话号码归在一起,然后再删除不符合要求的电话号码

#include <bits/stdc++.h>
using namespace std;
struct per
{
    char name[15];
    int num;
    int c[220];
    char tel[220][12];
};
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        per a[n+10];
        for(int i=0;i<n;i++)
        {
            scanf("%s",a[i].name);
            scanf("%d",&a[i].num);
            for(int j=0;j<a[i].num;j++)
            {
                scanf("%s",a[i].tel[j]);
            }
        }
        //printf("dddddd\n");
//        for(int i=0;i<n;i++)
//        {
//            printf("%s",a[i].name);
//            printf(" %d",a[i].num);
//            for(int j=0;j<a[i].num;j++)
//            {
//                printf(" %s",a[i].tel[j]);
//            }
//            printf("\n");
//        }
        per ans[n+10];
        int q=0;
        int b[25];
        memset(b,1,sizeof(b));
        for(int i=0;i<n;i++)
        {
            if(b[i])
            {
                b[i]=0;
                strcpy(ans[q].name,a[i].name);
                ans[q].num=a[i].num;
                for(int  j=0;j<ans[q].num;j++)
                    strcpy(ans[q].tel[j],a[i].tel[j]);
                for(int j=i+1;j<n;j++)
                {
                    if(strcmp(ans[q].name,a[j].name)==0)
                    {
                        b[j]=0;
                        for(int k=0;k<a[j].num;k++)
                        {
                            strcpy(ans[q].tel[ans[q].num++],a[j].tel[k]);
                        }
                    }
                }
                memset(ans[q].c,1,sizeof(ans[q].c));
                for(int j=0;j<ans[q].num;j++)
                {
                    if(ans[q].c[j])
                    {
                        for(int k=0;k<ans[q].num;k++)
                        {
                            if(k==j||ans[q].c[k]==0)
                                continue;
                            int l1=strlen(ans[q].tel[j]);
                            int l2=strlen(ans[q].tel[k]);
                            if(l2>l1)
                                continue;
                            int flag=1;
                            for(int l=l2-1;l>=0;l--,l1--)
                            {
                                if(ans[q].tel[k][l]!=ans[q].tel[j][l1-1])
                                {
                                    flag=0;
                                    break;
                                }
                            }
                            if(flag)
                                ans[q].c[k]=0;
                        //printf("----------\n");
                        }
                    }
                }
                q++;
            }
        }
//        printf("--------\n");
//        for(int i=0;i<q;i++)
//        {
//            printf("%s",ans[i].name);
//            printf(" %d",ans[i].num);
//            for(int j=0;j<ans[i].num;j++)
//            {
//                printf(" %s",ans[i].tel[j]);
//            }
//            printf("\n");
//        }
//        printf("-------\n");
        printf("%d\n",q);
        for(int i=0;i<q;i++)
        {
            //printf("%d\n",ans[i].num);
            printf("%s",ans[i].name);
            int m=0;
            for(int j=0;j<ans[i].num;j++)
            {
                if(ans[i].c[j])
                    m++;
            }
            printf(" %d",m);
            for(int j=0;j<ans[i].num;j++)
            {
                if(ans[i].c[j])
                {
                    printf(" %s",ans[i].tel[j]);
                }
            }
            printf("\n");
        }
    }
    return 0;
}


### 回答1: #include <stdio.h> int main(){ //定义一个3*3的数组用来存储棋盘 int board[3][3]={0}; int x,y,turn=1; while (1){ //打印棋盘 printf("当前棋盘:\n"); for (x=0;x<3;x++){ for (y=0;y<3;y++){ printf("%d ",board[x][y]); } printf("\n"); } //根据turn的值来判断谁轮到落子 if (turn==1){ printf("轮到X落子,请输入落子的位置(x y):"); }else { printf("轮到O落子,请输入落子的位置(x y):"); } scanf("%d %d",&x,&y); //将落子位置的值设置为对应的值 board[x][y] = turn; //改变轮到谁落子 turn = -turn; //判断谁赢了 if (board[0][0]==board[1][1] && board[1][1]==board[2][2] && board[2][2]!=0){ printf("游戏结束,获胜者是%c\n",board[0][0]==1?'X':'O'); break; } if (board[2][0]==board[1][1] && board[1][1]==board[0][2] && board[0][2]!=0){ printf("游戏结束,获胜者是%c\n",board[2][0]==1?'X':'O'); break; } for (x=0;x<3;x++){ if (board[x][0]==board[x][1] && board[x][1]==board[x][2] && board[x][2]!=0){ printf("游戏结束,获胜者是%c\n", board[x][0] == 1 ? 'X' : 'O'); break; } if (board[0][x]==board[1][x] && board[1][x]==board[2][x] && board[2][x]!=0){ printf("游戏结束,获胜者是%c\n", board[0][x] == 1 ? 'X' : 'O'); break; } } } return 0; } ### 回答2: 为了回答这个问题,需要提供题目的具体要求和规则。由于提供的信息不够具体,无法为您提供准确的代码。但是,我可以给您一个简单的Tic-tac-toe游戏的示例代码,供您参考: ```c #include <stdio.h> #include <stdbool.h> // 判断游戏是否结束 bool isGameOver(char board[][3]) { // 判断每行是否有3个相同的棋子 for(int i = 0; i < 3; i++) { if(board[i][0] != '.' && board[i][0] == board[i][1] && board[i][0] == board[i][2]) { return true; } } // 判断每列是否有3个相同的棋子 for(int i = 0; i < 3; i++) { if(board[0][i] != '.' && board[0][i] == board[1][i] && board[0][i] == board[2][i]) { return true; } } // 判断对角线是否有3个相同的棋子 if(board[0][0] != '.' && board[0][0] == board[1][1] && board[0][0] == board[2][2]) { return true; } if(board[0][2] != '.' && board[0][2] == board[1][1] && board[0][2] == board[2][0]) { return true; } return false; } // 输出棋盘 void printBoard(char board[][3]) { for(int i = 0; i < 3; i++) { for(int j = 0; j < 3; j++) { printf("%c ", board[i][j]); } printf("\n"); } } int main() { char board[3][3]; // 初始化棋盘 for(int i = 0; i < 3; i++) { for(int j = 0; j < 3; j++) { board[i][j] = '.'; } } int player = 1; // 玩家1先下 int row, col; while(true) { printf("Player %d's turn:\n", player); printf("Row: "); scanf("%d", &row); printf("Column: "); scanf("%d", &col); // 判断输入是否合法 if(row < 0 || row >= 3 || col < 0 || col >= 3 || board[row][col] != '.') { printf("Invalid move. Try again.\n"); continue; } // 下棋 board[row][col] = (player == 1) ? 'X' : 'O'; // 输出棋盘 printBoard(board); // 判断游戏是否结束 if(isGameOver(board)) { printf("Player %d wins!\n", player); break; } // 切换玩家 player = (player == 1) ? 2 : 1; } return 0; } ``` 这段代码实现了一个简单的命令行下的Tic-tac-toe游戏。玩家1使用'X'棋子,玩家2使用'O'棋子。玩家依次输入行和列,下棋后更新棋盘,并判断游戏是否结束。当游戏结束时,会输出获胜者并结束游戏。 ### 回答3: 题目要求实现一个井字棋游戏的判断胜负函数。给定一个3x3的井字棋棋盘,用C语言编写一个函数,判断当前是否存在某个玩家获胜或者平局。 题目要求代码中定义一个3x3的字符数组board来表示棋盘,其中 'X' 表示玩家1在该位置放置了一个棋子, 'O' 表示玩家2在该位置放置了一个棋子, '.' 表示该位置没有棋子。 下面是实现此题的C语言代码: ```c #include <stdio.h> #include <stdbool.h> // 用于使用bool类型 bool checkWin(char board[3][3]) { // 检查每一行是否有获胜的情况 for (int row = 0; row < 3; row++) { if (board[row][0] == board[row][1] && board[row][1] == board[row][2] && board[row][0] != '.') { return true; } } // 检查每一列是否有获胜的情况 for (int col = 0; col < 3; col++) { if (board[0][col] == board[1][col] && board[1][col] == board[2][col] && board[0][col] != '.') { return true; } } // 检查对角线是否有获胜的情况 if ((board[0][0] == board[1][1] && board[1][1] == board[2][2] && board[0][0] != '.') || (board[0][2] == board[1][1] && board[1][1] == board[2][0] && board[0][2] != '.')) { return true; } return false; // 没有获胜的情况 } int main() { char board[3][3]; // 存储棋盘状态 // 读取棋盘状态 for (int i = 0; i < 3; i++) { scanf("%s", board[i]); } // 调用检查胜负的函数,并输出结果 if (checkWin(board)) { printf("YES\n"); } else { printf("NO\n"); } return 0; } ``` 这个程序中定义了一个函数checkWin,用于检查是否有玩家获胜。遍历棋盘的每一行、每一列和对角线,判断是否有连续相同的字符且不为'.',如果有,则返回true;否则返回false。 在主函数main中,首先定义一个3x3的字符数组board,然后通过循环从标准输入中读取棋盘状态。接着调用checkWin函数进行胜负判断,并根据结果输出"YES"或者"NO"。最后返回0表示程序正常结束。 请注意,该代码只包含了检查胜负的功能,并没有包含其他如用户输入、判断平局等功能。如果需要完整的游戏代码,请告知具体要求。
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