生命游戏 蓝桥真题

直接模拟细胞繁殖与死亡的过程 打表记录下每次变化后的活细胞数量 发现有长度为60的循环节 即每60个数加10

 

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e2+10;

char ch[2*maxn][maxn][maxn];
int q=200,n=100,m=100;

void init()
{
    char pre[]="..........";
    int i,cnt;
    for(i=0;i<45;i++){
        strcpy(ch[0][i],pre);
        cnt=9;
        while(cnt--){
            strcat(ch[0][i],pre);
        }
    }

    for(i=45;i<55;i++){
        strcpy(ch[0][i],pre);
        cnt=2;
        while(cnt--){
            strcat(ch[0][i],pre);
        }
    }
    strcat(ch[0][45],"........................................");
    strcat(ch[0][46],"..........................X.............");
    strcat(ch[0][47],"........................X.X.............");
    strcat(ch[0][48],"..............XX......XX............XX..");
    strcat(ch[0][49],".............X...X....XX............XX..");
    strcat(ch[0][50],"..XX........X.....X...XX................");
    strcat(ch[0][51],"..XX........X...X.XX....X.X.............");
    strcat(ch[0][52],"............X.....X.......X.............");
    strcat(ch[0][53],".............X...X......................");
    strcat(ch[0][54],"..............XX........................");
    for(i=45;i<55;i++){
        cnt=3;
        while(cnt--){
            strcat(ch[0][i],pre);
        }
    }

    for(i=55;i<100;i++){
        strcpy(ch[0][i],pre);
        cnt=9;
        while(cnt--){
            strcat(ch[0][i],pre);
        }
    }
}

void solve()
{
    int next[8][2]={0,-1,-1,0,0,1,1,0,-1,-1,-1,1,1,1,1,-1};
    int i,j,k,l,tx,ty,cnt;
    for(i=1;i<=q;i++){
        for(j=0;j<n;j++){
            for(k=0;k<m;k++){
                cnt=0;
                for(l=0;l<8;l++){
                    tx=j+next[l][0],ty=k+next[l][1];
                    if(0<=tx&&tx<n&&0<=ty&&ty<m&&ch[i-1][tx][ty]=='X') cnt++;
                }
                if(ch[i-1][j][k]=='X'){
                    if(cnt==2||cnt==3) ch[i][j][k]='X';
                    else ch[i][j][k]='.';
                }
                else{
                    if(cnt==3) ch[i][j][k]='X';
                    else ch[i][j][k]='.';
                }
            }
        }
    }
}

void show()
{
    int i,j,k,cnt;
    for(i=1;i<=q;i++){
        cnt=0;
        for(j=0;j<n;j++){
            for(k=0;k<m;k++){
                printf("%c",ch[i][j][k]);
                if(ch[i][j][k]=='X') cnt++;
            }
            printf("\n");
        }
        printf("********************%d********************\n",cnt);
    }
}

bool judge(int a,int b)
{
    int res,i,j;
    res=1;
    for(i=0;i<n;i++){
        for(j=0;j<m;j++){
            if(ch[a][i][j]!=ch[b][i][j]){
                res=0;
                break;
            }
        }
        if(!res) break;
    }
    return res;
}

void getloop()
{
    int i,j;
    for(i=1;i<=q;i++){
        for(j=i+1;j<=q;j++){
            if(judge(i,j)){
                printf("%d %d\n",i,j);
            }
        }
    }
}

void getnum()
{
    int val[2*maxn];
    int i,j,k;
    for(i=1;i<=q;i++){
        val[i]=0;
        for(j=0;j<n;j++){
            for(k=0;k<m;k++){
                if(ch[i][j][k]=='X') val[i]++;
            }
        }
    }
    for(i=1;i<=q;i++){
        printf("%d",val[i]);
        if(i%60==0) printf("\n");
        else printf(" ");
    }
    printf("\n");
    printf("***%d***\n",val[40]);
}

int main()
{
    init();
    solve();
    //show();
    //getloop();
    getnum();
    return 0;
}

/*
........................................
..........................X.............
........................X.X.............
..............XX......XX............XX..
.............X...X....XX............XX..
..XX........X.....X...XX................
..XX........X...X.XX....X.X.............
............X.....X.......X.............
.............X...X......................
..............XX........................
*/