/ Vijos / 题库 / 扫雷游戏 /

题解

54 条题解

  • 4
    @ 2017-10-04 00:17:00

    直接覆盖上去玩了。
    #include <iostream>
    using namespace std;

    char a[105][105];
    int b[105][105];
    int n,m;
    char temp;

    int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
    for(int j=1;j<=m;j++){
    temp=getchar();
    if((temp=='*')||(temp=='?')){
    a[i][j]=temp;
    }else{
    a[i][j]=getchar();
    }
    }

    }
    for(int i=1;i<=n;i++){
    for(int j=1;j<=n;j++){
    if(a[i][j]=='*'){
    for(int x=i-1;x<=i+1;x++){
    for(int y=j-1;y<=j+1;y++){
    b[x][y]++;
    }
    }
    }
    }
    }
    for(int i=1;i<=n;i++){
    for(int j=1;j<=m;j++){
    if(a[i][j]=='*'){
    cout<<'*';
    }else{
    cout<<b[i][j];
    }
    }
    cout<<endl;
    }
    return 0;
    }

  • 2
    @ 2016-11-16 16:01:03

    #include<iostream>
    #include<string.h>
    using namespace std;
    char a[105][105];
    int m,n,b[105][105];
    int c[16]={0,1,0,-1,1,0,-1,0,1,1,1,-1,-1,1,-1,-1};
    int main()
    {
    //freopen("b.in","r",stdin);
    //freopen("b.out","w",stdout);
    cin>>n>>m;
    memset(b,0,sizeof(b));
    for(int i=0;i<n;i++)
    {
    for(int j=0;j<m;j++)cin>>a[i][j];
    }
    for(int i=0;i<n;i++)
    {
    for(int j=0;j<m;j++)
    {
    if(a[i][j]=='*')
    {
    for(int k=0;k<16;k+=2)if(i+c[k]>=0&&i+c[k]<n&&j+c[k+1]>=0&&j+c[k+1]<m)b[i+c[k]][j+c[k+1]]++;
    }
    }
    }
    for(int i=0;i<n;i++)
    {
    for(int j=0;j<m;j++)
    {
    if(a[i][j]=='?')a[i][j]='0'+b[i][j];
    }
    }
    for(int i=0;i<n;i++)
    {
    for(int j=0;j<m;j++)cout<<a[i][j];
    cout<<endl;
    }
    return 0;
    }

  • 1
    @ 2022-07-20 13:44:53 
    #include <iostream>
using namespace std;

char a[105][105];
int b[105][105];
int n,m;
char temp;

int main(){
cin>>n>>m;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
temp=getchar();
if((temp=='*')||(temp=='?')){
a[i][j]=temp;
}else{
a[i][j]=getchar();
}
}

}
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(a[i][j]=='*'){
for(int x=i-1;x<=i+1;x++){
for(int y=j-1;y<=j+1;y++){
b[x][y]++;
}
}
}
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(a[i][j]=='*'){
cout<<'*';
}else{
cout<<b[i][j];
}
}
cout<<endl;
}
return 0;
}
  • 1
    @ 2021-08-29 16:47:18 
    #include<bits/stdc++.h>
using namespace std;

bool a[105][105];
int main()
{
    memset(a,0,sizeof(a));
    int n,m;
    char tmp;
    cin>>n>>m; 
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++) 
        {
            cin>>tmp; 
            if(tmp=='*') a[i][j]=1; 
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]==1) printf("*");
            else
            {
                printf("%d",a[i+1][j+1]+a[i+1][j-1]+a[i+1][j]+a[i][j+1]+a[i][j-1]+a[i-1][j+1]+a[i-1][j]+a[i-1][j-1]);
            }
        }
        printf("\n");
    }
    return 0;
}
  • 1
    @ 2021-02-26 08:41:54 
    #include<bits/stdc++.h>
using  namespace  std;
int n,k,a[110][110],ans,t,m,l=1;
char b[110][110];
int  main()
{
//freopen("mine.in","r",stdin);
 //freopen("mine.out","w",stdout);
    cin>>n>>m;
    for(int i=1;i<=n;i=i+1)
     for(int j=1;j<=m;j++)
        cin>>b[i][j];
         
    for(int i=1;i<=n;i++)
    {
      for(int j=1;j<=m;j++)
      {
         if(b[i][j]!='*')
         {
            if(b[i-1][j-1]=='*')a[i][j]++;
            if(b[i][j-1]=='*')a[i][j]++;
            if(b[i+1][j-1]=='*')a[i][j]++;
            if(b[i+1][j]=='*')a[i][j]++;
            if(b[i+1][j+1]=='*')a[i][j]++;
            if(b[i][j+1]=='*')a[i][j]++;
            if(b[i-1][j+1]=='*')a[i][j]++;
            if(b[i-1][j]=='*')a[i][j]++;
         }
         else a[i][j]=-1;
      }
    }
    for(int i=1;i<=n;i++)
    {
      for(int j=1;j<=m;j++)
      {
        if(a[i][j]!=-1)cout<<a[i][j];
        else cout<<"*";
      }
      cout<<endl;
    }
       
    return 0;
}
  • 1
    @ 2018-08-20 09:23:54 
    /*
水题啊
直接乱搞不就好了
对于每一个点
我们来直接对于每一个点
就直接搜索一下八个方向个数就好
水题Orz
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAXN=105;
int n,m;
int a[MAXN][MAXN];
int zx[8][2]={{-1,0},{-1,-1},{1,1},{0,-1},{1,0},{-1,1},{1,-1},{0,1}};

void init()
{
    char ch;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            cin>>ch;
            if(ch=='*')
                a[i][j]=-1;
        }
}

int gets(int x,int y)
{
    int ans=0;
    for(int i=0;i<8;i++)
    {
        int newx=x+zx[i][0];
        int newy=y+zx[i][1];
        if(newx>n||newy>m||newx<1||newy<1)
            continue;
        if(a[newx][newy]==-1)
            ans++;
    }
    return ans;
}

int main()
{
    init();
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(a[i][j]!=-1)
                a[i][j]=gets(i,j);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
            if(a[i][j]==-1)
                cout<<'*';
            else
                cout<<a[i][j];
        cout<<endl;
    }
}
  • 1
    @ 2017-10-02 21:45:56

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    char a[105][105];
    int b[105][105];
    int main()
    {
    //freopen("game.in","r",stdin);
    //freopen("game.out","w",stdout);
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    {
    cin>>a[i][j];
    if(a[i][j]=='*')
    {
    b[i][j]=10;
    b[i+1][j]++;
    b[i][j+1]++;
    b[i+1][j+1]++;
    b[i+1][j-1]++;
    b[i][j-1]++;
    b[i-1][j-1]++;
    b[i-1][j]++;
    b[i-1][j+1]++;
    }
    }
    for(int i=1;i<=n;i++)
    {
    for(int j=1;j<=m;j++)
    {
    if(b[i][j]<10)
    cout<<b[i][j];
    else
    cout<<"*";
    }
    cout<<endl;
    }
    return 0;
    }

  • 1
    @ 2017-01-23 00:50:25

    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    char s[1001][1001];
    int main(){
    int i,j,k,n,m;
    scanf("%d%d\n",&n,&m);
    for(i=1;i<=n;i++){
    for(j=1;j<=m;j++)
    scanf("%c",&s[i][j]);
    if(i!=n)scanf("\n");
    }
    for(i=1;i<=n;i++){
    for(j=1;j<=m;j++)
    if(s[i][j]=='*')printf("*");
    else{
    int la=0;
    if(s[i-1][j]=='*')la++;
    if(s[i-1][j-1]=='*')la++;
    if(s[i-1][j+1]=='*')la++;
    if(s[i+1][j]=='*')la++;
    if(s[i+1][j-1]=='*')la++;
    if(s[i+1][j+1]=='*')la++;
    if(s[i][j-1]=='*')la++;
    if(s[i][j+1]=='*')la++;
    printf("%d",la);
    }
    printf("\n");
    }
    return 0;
    }

  • 1
    @ 2016-11-11 22:03:48 
    var
  n,m,i,j:longint;
  a,b:array[1..100,1..100] of char;
begin
  fillchar(b,sizeof(b),'0');
  readln(n,m);
  for i:=1 to n do
    begin
      for j:=1 to m do
        read(a[i,j]);
      readln;
    end;
  for i:=1 to n do
    for j:=1 to m do
      begin
        if a[i,j] = '?' then
          begin
            if j <> 1 then
              if a[i,j-1] = '*' then
                b[i,j]:=chr(ord(b[i,j])+1); //left
            if (i <> 1) and (j <> 1) then
              if a[i-1,j-1] = '*' then
                b[i,j]:=chr(ord(b[i,j])+1); //left up
            if i <> 1 then
              if a[i-1,j] = '*' then
                b[i,j]:=chr(ord(b[i,j])+1); //up
            if (i <> 1) and (j <> m) then
              if a[i-1,j+1] = '*' then
                b[i,j]:=chr(ord(b[i,j])+1); //right up
            if j <> m then
              if a[i,j+1] = '*' then
                b[i,j]:=chr(ord(b[i,j])+1); //right
            if (i <> n) and (j <> m) then
              if a[i+1,j+1] = '*' then
                b[i,j]:=chr(ord(b[i,j])+1); //right down
            if i <> n then
              if a[i+1,j] = '*' then
                b[i,j]:=chr(ord(b[i,j])+1); //down
            if (i <> n) and (j <> 1) then
              if a[i+1,j-1] = '*' then
                b[i,j]:=chr(ord(b[i,j])+1); //left down
          end;
       if a[i,j] = '*' then
         b[i,j]:='*';
      end;
  for i:=1 to n do
    begin
      for j:=1 to m do
        write(b[i,j]);
      writeln;
    end;
end.
  • 1
    @ 2016-11-02 13:14:31

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #define N 101
    using namespace std;
    char a[N][N];
    int tot;
    int n,m;
    int search(int x,int y )
    {
    if(x-1>=1&&y-1>=1&&a[x-1][y-1]=='*')
    tot++;
    if(x+1<=n&&y+1<=m&&a[x+1][y+1]=='*')
    tot++;
    if(x-1>=1&&a[x-1][y]=='*')
    tot++;
    if(x-1>=1&&y+1<=m&&a[x-1][y+1]=='*')
    tot++;
    if(y-1>=1&&a[x][y-1]=='*')
    tot++;
    if(y+1<=m&&a[x][y+1]=='*')
    tot++;
    if(x+1<=n&&y-1>=1&&a[x+1][y-1]=='*')
    tot++;
    if(x+1<=n&&a[x+1][y]=='*')
    tot++;
    return tot;
    }
    int main()
    {
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    {
    cin>>a[i][j];
    }
    for(int i=1;i<=n;i++)
    {
    for(int j=1;j<=m;j++)
    {
    if(a[i][j] == '?')
    {
    tot=0;
    printf("%d",search(i,j));
    }
    if(a[i][j]=='*')
    printf("*");
    }
    cout<<endl;
    }
    return 0;
    }

  • 1
    @ 2016-10-28 19:50:00

    #include <bits/stdc++.h>
    using namespace std;
    char a[105][105];
    int b[105][105];
    bool c[105][105];
    int main()
    {
    int i,j,k,l,n,m;
    cin>>n>>m;
    memset(a,1,sizeof(c));
    for(i=1;i<=n;i++) {
    for(j=1;j<=m;j++) {
    cin>>a[i][j];
    if(a[i][j]=='*') {
    b[i][j]=-1000000;
    if(b[i+1][j]!=-1000000) b[i+1][j]++;
    if(b[i-1][j]!=-1000000) b[i-1][j]++;
    if(b[i][j+1]!=-1000000) b[i][j+1]++;
    if(b[i][j-1]!=-1000000) b[i][j-1]++;
    if(b[i+1][j+1]!=-1000000) b[i+1][j+1]++;
    if(b[i-1][j+1]!=-1000000) b[i-1][j+1]++;
    if(b[i+1][j-1]!=-1000000) b[i+1][j-1]++;
    if(b[i-1][j-1]!=-1000000) b[i-1][j-1]++;
    }
    }
    }
    for(i=1;i<=n;i++){
    for(j=1;j<=m;j++){
    if(b[i][j]!=-1000000) cout<<b[i][j];
    else cout<<"*";
    }
    cout<<endl;
    }
    return 0;
    }
    测试数据 #0: Accepted, time = 15 ms, mem = 624 KiB, score = 10

    测试数据 #1: Accepted, time = 0 ms, mem = 624 KiB, score = 10

    测试数据 #2: Accepted, time = 15 ms, mem = 620 KiB, score = 10

    测试数据 #3: Accepted, time = 0 ms, mem = 628 KiB, score = 10

    测试数据 #4: Accepted, time = 0 ms, mem = 624 KiB, score = 10

    测试数据 #5: Accepted, time = 15 ms, mem = 624 KiB, score = 10

    测试数据 #6: Accepted, time = 0 ms, mem = 624 KiB, score = 10

    测试数据 #7: Accepted, time = 15 ms, mem = 624 KiB, score = 10

    测试数据 #8: Accepted, time = 0 ms, mem = 624 KiB, score = 10

    测试数据 #9: Accepted, time = 15 ms, mem = 624 KiB, score = 10

    Accepted, time = 75 ms, mem = 628 KiB, score = 100

  • 1
    @ 2016-10-14 13:58:57

    program ex2;
    var i,j,l,t,m,n:integer;
    s:array[1..102,1..102] of char;
    a:array[1..102,1..102] of integer;
    b:array[1..102,1..102] of boolean;
    begin
    readln(n,m);
    for i:=1 to n+2 do
    for j:=1 to m+2 do
    begin
    s[i,j]:='?';
    a[i,j]:=0;
    end;
    for i:=2 to n+1 do
    begin
    for j:=2 to m+1 do
    begin
    read(s[i,j]);
    if s[i,j]='?' then b[i,j]:=true
    else b[i,j]:=false;
    end;
    readln;
    end;
    for i:=2 to n+1 do
    for j:=2 to m+1 do
    for l:=i-1 to i+1 do
    for t:=j-1 to j+1 do
    if s[l,t]='*' then a[i,j]:=a[i,j]+1;
    for i:=2 to n+1 do
    begin
    for j:=2 to m+1 do
    if b[i,j] then write(a[i,j])
    else write('*');
    writeln;
    end;
    end.

  • 1
    @ 2016-10-02 11:21:25

    #include <iostream>
    using namespace std;
    char a[107][107],b[107][107];
    int x[10]={1,1,1,0,0,0,-1,-1,-1},y[10]={1,0,-1,1,0,-1,1,0,-1},cur;
    int main()
    {
    int m,n;
    cin>>n>>m;
    for(int i=0;i<=n+2;i++)
    for(int j=0;j<=m+2;j++)
    a[i][j]='?';
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    cin>>a[i][j];
    for(int i=1;i<=n;i++)
    {
    for(int j=1;j<=m;j++)
    {
    cur=0;
    if(a[i][j]=='*') cout<<"*";
    else
    {
    for(int k=0;k<=8;k++)
    if(a[i+x[k]][j+y[k]]=='*')
    cur++;
    cout<<cur;
    }
    }
    cout<<endl;
    }
    system ("pause");
    return 0;
    }

  • 1
    @ 2016-09-04 19:47:34

    var a:array[0..100,0..100]of char;
    n,m,t,i,j:longint;
    begin
    readln(n,m);
    for i:=1 to n do begin
    for j:=1 to m do
    read(a[i,j]);
    readln;
    end;
    for i:=1 to n do begin
    for j:=1 to m do begin
    t:=0;
    if a[i,j]='*' then write('*')
    else begin
    if a[i-1,j]='*' then inc(t);
    if a[i-1,j-1]='*' then inc(t);
    if a[i-1,j+1]='*' then inc(t);
    if a[i,j-1]='*' then inc(t);
    if a[i,j+1]='*' then inc(t);
    if a[i+1,j]='*' then inc(t);
    if a[i+1,j-1]='*' then inc(t);
    if a[i+1,j+1]='*' then inc(t);
    write(t);
    end;
    end;
    writeln;
    end;
    end.
    比赛现场打的，好傻啊。。。

  • 1
    @ 2015-12-21 15:46:40

    记录信息
    评测状态 Accepted
    题目 P1975 扫雷游戏
    递交时间 2015-12-20 16:50:50
    代码语言 Pascal
    评测机 VijosEx
    消耗时间 308 ms
    消耗内存 656 KiB
    评测时间 2015-12-20 16:50:51

    评测结果
    编译成功

    Free Pascal Compiler version 2.6.2 [2013/02/12] for i386
    Copyright (c) 1993-2012 by Florian Klaempfl and others
    Target OS: Win32 for i386
    Compiling foo.pas
    foo.pas(4,1) Note: Local variable "ip" not used
    foo.pas(4,4) Note: Local variable "op" not used
    Linking foo.exe
    33 lines compiled, 0.1 sec , 28864 bytes code, 1628 bytes data
    2 note(s) issued
    测试数据 #0: Accepted, time = 46 ms, mem = 652 KiB, score = 10

    测试数据 #1: Accepted, time = 31 ms, mem = 652 KiB, score = 10

    测试数据 #2: Accepted, time = 31 ms, mem = 652 KiB, score = 10

    测试数据 #3: Accepted, time = 31 ms, mem = 652 KiB, score = 10

    测试数据 #4: Accepted, time = 31 ms, mem = 652 KiB, score = 10

    测试数据 #5: Accepted, time = 46 ms, mem = 652 KiB, score = 10

    测试数据 #6: Accepted, time = 31 ms, mem = 652 KiB, score = 10

    测试数据 #7: Accepted, time = 31 ms, mem = 652 KiB, score = 10

    测试数据 #8: Accepted, time = 15 ms, mem = 656 KiB, score = 10

    测试数据 #9: Accepted, time = 15 ms, mem = 652 KiB, score = 10

    Accepted, time = 308 ms, mem = 656 KiB, score = 100

    代码
    var
    l:array[1..200,1..200]of char;
    i,j,k,m,n:longint;
    ip,op:text;
    begin

    readln(m,n);
    fillchar(l,sizeof(l),'0');
    for i:=2 to m+1 do begin//这里从位置2，2开始读是为了避免下标越界报错
    for j:=2 to n+1 do read(l[i,j]);
    readln;//注意这里一定要写，不写会读错的
    end;
    for i:=2 to m+1 do begin
    for j:=2 to n+1 do begin
    if l[i,j]='*' then
    write('*');//如果是雷直接输出
    if l[i,j]='?' then begin
    k:=0;
    if l[i,j+1]='*' then inc(k);//if语句判断是否加一
    if l[i,j-1]='*' then inc(k);//同上
    if l[i+1,j]='*' then inc(k);//同上
    if l[i-1,j]='*' then inc(k);//同上
    if l[i+1,j+1]='*' then inc(k);//同上
    if l[i+1,j-1]='*' then inc(k);//同上
    if l[i-1,j+1]='*' then inc(k);//同上
    if l[i-1,j-1]='*' then inc(k);//同上
    write(k);
    end;
    end;
    writeln;//这边不加的话就变成一行了
    end;

    end.

  • 0
    @ 2024-09-08 12:27:15 
    #include<iostream>
using namespace std;
char a[110][110];
int q[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
int main(){
    ios::sync_with_stdio(false);
    int m, n, cnt = 0;
    cin >> m >> n;
    for(int i = m+1; i >= 0; i--){
        for(int j = n+1; j >= 0; j--)a[i][j] = '?';
    }
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j++)cin >> a[i][j];
    }
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j++){
            if(a[i][j] == '?'){
                for(int k = 0; k < 8; k++){
                    if(a[i+q[k][0]][j+q[k][1]] == '*')cnt++;
                }
                cout << cnt;
                cnt = 0;
            }else cout << '*';
        }
        cout << endl;
    }
    return 0;
}
  • 0
    @ 2018-06-21 15:22:59

    #include <iostream>
    #include <cstdio>
    using namespace std;
    int main()
    {
    int n, m;
    cin>>n>>m;
    int i, j, count = 0;;
    char array[n][m];
    for(i = 0; i < n; i++) {
    for(j = 0; j < m; j++) {
    cin>>array[i][j];
    if(array[i][j] == '*')
    count++;
    }
    }
    if(count == n * m)
    {
    for(i = 0; i < n; i++) {
    for(j = 0; j < m; j++) {
    cout<<array[i][j];
    }
    cout<<endl;
    }
    }
    else
    {
    for(i = 0; i < n; i++) {
    for(j = 0; j < m; j++) {
    if(array[i][j] == '?')
    array[i][j] = '0';
    }
    }
    for(i = 0; i < n; i++) {
    for(j = 0; j < m; j++) {
    if(array[i][j] == '0')
    {
    if(array[i][j+1] == '*' && i < n && j + 1 < m && i >= 0 && j + 1 >= 0)
    array[i][j]++;
    if(array[i][j-1] == '*' && i < n && j - 1 < m && i >= 0 && j - 1 >= 0)
    array[i][j]++;
    if(array[i+1][j] == '*' && i + 1 < n && j < m && i + 1 >= 0 && j >= 0)
    array[i][j]++;
    if(array[i-1][j] == '*' && i - 1 < n && j < m && i - 1 >= 0 && j >= 0)
    array[i][j]++;
    if(array[i+1][j+1] == '*' && i + 1 < n && j + 1 < m && i + 1 >= 0 && j + 1 >= 0)
    array[i][j]++;
    if(array[i+1][j-1] == '*' && i + 1 < n && j - 1 < m && i + 1 >= 0 && j - 1 >= 0)
    array[i][j]++;
    if(array[i-1][j+1] == '*' && i - 1 < n && j + 1 < m && i - 1 >= 0 && j + 1 >= 0)
    array[i][j]++;
    if(array[i-1][j-1] == '*' && i - 1 < n && j - 1 < m && i - 1 >= 0 && j - 1 >= 0)
    array[i][j]++;
    }
    }
    }

    for(i = 0; i < n; i++) {
    for(j = 0; j < m; j++) {
    cout<<array[i][j];
    }
    cout<<endl;
    }
    }
    return 0;
    }

  • 0
    @ 2018-06-19 19:36:36 
    #include<iostream>
using namespace std;
int main()
{
    int n,m,map[102][102]={0};
    char tmp;
    cin>>n>>m;
    for (int h=1;h<=n;h++)
        for (int l=1;l<=m;l++)
        {
            cin>>tmp;
            if (tmp=='*')
                map[h][l]=1;
        }
    for (int h=1;h<=n;h++)
    {
        for (int l=1;l<=m;l++)
        {
            if (map[h][l]==1)
                cout<<'*';
            else
            {
                int tot=0;
                for (int lh=h-1;lh<=h+1;lh++)
                    for (int ll=l-1;ll<=l+1;ll++)
                        if (map[lh][ll]==1)
                            tot++;
                cout<<tot;
            }               
        }
        cout<<endl;
    }
    
    return 0;
}
  • 0
    @ 2017-06-04 13:01:17

    #include <cstring>
    #include <iostream>
    using namespace std;
    char a[110][110];
    int m,n;
    int x[9]={0,1,1,1,0,0,-1,-1,-1};
    int y[9]={0,1,0,-1,1,-1,1,0,-1};
    int main()
    {
    cin>>m>>n;
    for(int i=1;i<=m;i++) cin>>a[i]+1;
    for(int i=1;i<=m;i++)
    {
    for(int j=1;j<=n;j++)
    if(a[i][j]=='*') cout<<"*";
    else
    {
    int flag=0;
    for(int t=1;t<=8;t++)
    flag+=(a[i+x[t]][j+y[t]]=='*');
    cout<<flag;
    }

    cout<<"\n";
    }
    return 0;
    }

  • 0
    @ 2016-08-25 12:37:45

    这里只是102和105的区别，QAQ………………………………………………………………

    编译成功

    测试数据 #0: RuntimeError, time = 15 ms, mem = 596 KiB, score = 0
    测试数据 #1: RuntimeError, time = 15 ms, mem = 596 KiB, score = 0
    测试数据 #2: RuntimeError, time = 0 ms, mem = 596 KiB, score = 0
    测试数据 #3: RuntimeError, time = 15 ms, mem = 596 KiB, score = 0
    测试数据 #4: RuntimeError, time = 15 ms, mem = 596 KiB, score = 0
    测试数据 #5: RuntimeError, time = 0 ms, mem = 596 KiB, score = 0
    测试数据 #6: RuntimeError, time = 0 ms, mem = 596 KiB, score = 0
    测试数据 #7: RuntimeError, time = 0 ms, mem = 600 KiB, score = 0
    测试数据 #8: RuntimeError, time = 0 ms, mem = 596 KiB, score = 0
    测试数据 #9: RuntimeError, time = 0 ms, mem = 596 KiB, score = 0
    RuntimeError, time = 60 ms, mem = 600 KiB, score = 0
    代码
    c++
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    using namespace std;
    int u[9]={-10,0,1,0,-1,1,1,-1,-1},w[9]={-10,1,0,-1,0,1,-1,1,-1};
    int main(void)
    {
    int a,b;
    cin>>a>>b;
    char maps[102][102];
    for(int i=1;i<=a;i++)
    for(int j=1;j<=b;j++)
    cin>>maps[i][j];
    int map[101][101];
    memset(map,0,sizeof(map));
    for(int i=1;i<=a;i++)
    for(int j=1;j<=b;j++)
    if(maps[i][j]=='*')map[i][j]=-1;
    for(int i=1;i<=a;i++)
    for(int j=1;j<=b;j++)
    if(map[i][j]!=-1)
    for(int k=1;k<=8;k++)
    {
    if(map[i+u[k]][j+w[k]]==-1)
    map[i][j]++;
    }
    for(int i=1;i<=a;i++)
    {
    for(int j=1;j<=b;j++)
    {
    if(map[i][j]!=-1)cout<<map[i][j];
    else cout<<"*";
    }
    cout<<endl;
    }
    return 0;
    }

    -----------哭泣的分割线--------------------

    编译成功
    测试数据 #0: Accepted, time = 15 ms, mem = 616 KiB, score = 10
    测试数据 #1: Accepted, time = 15 ms, mem = 616 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 616 KiB, score = 10
    测试数据 #3: Accepted, time = 15 ms, mem = 620 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 620 KiB, score = 10
    测试数据 #5: Accepted, time = 0 ms, mem = 612 KiB, score = 10
    测试数据 #6: Accepted, time = 15 ms, mem = 616 KiB, score = 10
    测试数据 #7: Accepted, time = 15 ms, mem = 616 KiB, score = 10
    测试数据 #8: Accepted, time = 0 ms, mem = 616 KiB, score = 10
    测试数据 #9: Accepted, time = 15 ms, mem = 612 KiB, score = 10
    Accepted, time = 90 ms, mem = 620 KiB, score = 100
    代码
    ```c++

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    using namespace std;
    int u[9]={-10,0,1,0,-1,1,1,-1,-1},w[9]={-10,1,0,-1,0,1,-1,1,-1};
    int main(void)
    {
    int a,b;
    cin>>a>>b;
    char maps[105][105];
    for(int i=1;i<=a;i++)
    for(int j=1;j<=b;j++)
    cin>>maps[i][j];
    int map[105][105];
    memset(map,0,sizeof(map));
    for(int i=1;i<=a;i++)
    for(int j=1;j<=b;j++)
    if(maps[i][j]=='*')map[i][j]=-1;
    for(int i=1;i<=a;i++)
    for(int j=1;j<=b;j++)
    if(map[i][j]!=-1)
    for(int k=1;k<=8;k++)
    {
    if(map[i+u[k]][j+w[k]]==-1)
    map[i][j]++;
    }
    for(int i=1;i<=a;i++)
    {
    for(int j=1;j<=b;j++)
    {
    if(map[i][j]!=-1)cout<<map[i][j];
    else cout<<"*";
    }
    cout<<endl;
    }
    return 0;
    }

    ```

扫雷游戏

信息

ID
1975
难度
4
分类
模拟 点击显示
标签
递交数
2825
已通过
1094
通过率
39%
被复制
23
上传者
  1. 关于
  2. 联系我们
  3. 隐私
  4. 服务条款
  5. 版权申诉
  6. Language
  1. © 2005 - 2023 Vijos.org
  2. uibuild-2f1065f
  3. 沪ICP备14040537号-1