2013-5-22 完美世界复赛第三场

做了前两场虽然有点恶心，但也没今天那么想吐嘈.... 反正是一题未A..不得不承认依旧很菜..

 

A题,死了命的提示结果错误....  显然已模拟题.封装 remove 与 maintain 然后print.. 各种情况都考虑,将其后台数据都输出比较.还是未找到错误点在哪里..

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

using namespace std;



const int N = 50;

char mp[N][N];

int n, m;

bool vis[N][N];



bool legal(int x,int y){

    if(x>=1&&x<=n&&y>=1&&y<=m)

        return true;

    return false;

}

void maintain(){



    for(int c = 1; c <= m; c++){



        for(int r = n; r > 1; r--){

            if( mp[r][c] == '0' ){

                int p = -1;        

                for(int t = r-1; t >= 1; t-- ){

                    if( mp[t][c] != '0' ){

                        p = t; break;    

                    }

                }

                if( p == -1 ) break;

                else swap( mp[r][c], mp[p][c] );

            }    

        }        

    }

}



bool find(int x,int y){

    bool flag = false;    

    // left 

    if( y >= 3 ){

        char ch = mp[x][y];

        if( (mp[x][y-1]==ch) && (mp[x][y-2]==ch) ){

            flag = true;    

            for(int i = y; i >= 1; i--)    

                if( mp[x][i] == ch ) vis[x][i] = true;

                else break;

        }

    }    

    // right

    if( y+2 <= m ){

        char ch = mp[x][y];

        if( (mp[x][y+1]==ch) && (mp[x][y+2]==ch) ){

            flag = true;

            for(int i = y; i <= m; i++)

                if( mp[x][i] == ch ) vis[x][i] = true;

                else break;

        }

    }    

    // up

    if( x >= 3 ){

        char ch = mp[x][y];

        if( (mp[x-1][y]==ch) && (mp[x-2][y]==ch) ){

            flag = true;

            for(int i = x; i >= 1; i--)

                if( mp[i][y] == ch ) vis[i][y] = true;

                else break;

        }

    }    

    // down

    if( x+2 <= n ){

        char ch = mp[x][y];

        if( (mp[x+1][y]==ch) && (mp[x+2][y]==ch) ){

            flag = true;

            for(int i = x; x <= n; i++)

                if( mp[i][y] == ch ) vis[i][y] = true;

                else    break;

        }

    }

    return flag;

}

bool remove(){

    memset(vis,0,sizeof(vis));

    bool flag = false;    

    for(int i = 1; i <= n; i++)

        for(int j = 1; j <= m; j++){

            if( (!vis[i][j]) && (mp[i][j]!='0') ){

                if( find(i,j) )

                    flag = true;

            }    

        }

    if( flag ){    

        for(int i = 1; i <= n; i++)

            for(int j = 1; j <= m; j++)

                if( vis[i][j] ) mp[i][j] = '0';

    }    

    return flag;

}

void print(){

    for(int i = 1; i <= n; i++){

        for(int j = 1; j <= m; j++)

            printf("%c", mp[i][j]);

        puts("");

    }

}



int main(){

    freopen("1.in","r",stdin);

    int T;

    scanf("%d", &T);

    while(T--){

        scanf("%d%d",&n,&m); 

        for(int i = 1; i <= n; i++)

            scanf("%s", mp[i]+1 );    

        

        maintain();

        while( remove() ){

        //    print();    

            maintain();

            print();    

        }            

    }

    return 0;

}

View Code

 

B题 . 魔方构造. 坑爹的是竟然还有N=2....  神一般的构造..看文库 http://wenku.baidu.com/view/02ace38ad0d233d4b14e693f.html   一般人都会..没看估计都不会.....

　　对于题目中 "对于每个N字图，每行输出N字图的一行，每行中的数字之间用一个或多个空格分开（注意对齐方式需要按最大的那个数字来对齐）", 真心想吐嘈... 这尼码是啥要求啊.所谓最大数字来对齐, 第一 是左还是右啊, 是 按最大的 n*n来, 还是当前一行的最大呢..  您好歹给点详细说明呗. 对于N=2这种无解的情况也没个说法,该如何处理.....

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

using namespace std;



const int N = 110;





void gao(int mp[N][N], int n, int c){

    int x = 0, y = n/2;

    mp[x][y] = 1 + c;    

    for(int top = 2; top <= n*n; top++){

        int x1 = (x-1+n)%n, y1 = (y+1+n)%n;

        if( mp[x1][y1] )

            x = (x+1+n)%n;    

        else x = x1, y = y1;        

        mp[x][y] = top + c;    

    }    

    

}

void kao(int mp[N][N], int n){

    if( n%4 == 0 ){

        for(int i = 0, top = 1; i < n; i++)

            for(int j = 0; j < n; j++)

                mp[i][j] = top++;

        for(int i = 0; i < n; i++){

            for(int j = 0; j < n-i; j++){

                if( (i==j)||(i+j==n-1) ) continue;

                swap( mp[i][j], mp[n-1-i][n-1-j] );

            }

        }

    }

}

int tmp[N][N];

int A[N][N], B[N][N], C[N][N], D[N][N];

int mp[N][N];



void cao(int n){

    int m = n/4;

    int k = 2*m+1;

    gao( A, k, 0 ); gao( B, k, k*k );

    gao( C, k, 2*k*k ); gao( D, k, 3*k*k );



    for(int i = 0; i < k; i++){

        int d = (i==k/2)?1:0;    

        for(int j = 0; j < m; j++){

            swap( A[i][j+d], D[i][j+d] );    

        }

    }

    for(int i = 0; i < k; i++){

        for(int j = 0; j < m-1; j++){

            swap( C[i][k-1-j], B[i][k-1-j] );    

        }    

    }    

    for(int i = 0; i < k; i++)

        for(int j = 0; j < k; j++){

            mp[i][j] = A[i][j];

            mp[i][j+3] = C[i][j];

            mp[i+3][j] = D[i][j];

            mp[i+3][j+3] = B[i][j];

                

        }

}

int n;



void print(int x){

    int la = 0, lb = 0, t = n*n;

    while( t ) la++, t /= 10;

    t = x;

    while( t ) lb++, t /= 10;

    printf("%d", x);    

    for(int i = lb; i < la; i++)

        printf(" ");

}

int main(){

    while( scanf("%d", &n), n ){

        memset( mp, 0, sizeof(mp));    

        

        if( n == 2 ){ continue; }    

        if( n&1 ) gao(mp, n, 0);

        else if( n%4 == 0 ) kao(mp, n);

        else cao( n );    

        for(int i = 0; i < n; i++){

            for(int j = 0; j < n; j++){

                if( j != 0 ) printf(" ");

                //printf("%d", mp[i][j] );    

                print( mp[i][j] );

            }

            puts("");    

        }

        puts("");    

    }    

    return 0;

}

View Code

 

C题, 不知道啥题..不过 岛娘A掉了.. Orz... 

 

总结, 完美世界真心坑..好想知道有多少人在参加这个........ -_-!!!