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;
}

 

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;
}

 

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........ -_-!!!