Timus 1219

#include <iostream>
#include <fstream>
using namespace std;

unsigned short counter_3D[26][26][26] = { 0 };
unsigned short counter_2D[26][26] = { 0 };
unsigned short counter_1D[26] = { 0 };
char ret[1000001] = { '\0' };
char seq[3] = { 0 };


int cnt=0,check[26] = { 0 }, check2D[26][26] = { 0 }, check3D[26][26][26] = { 0 };
char test[1000001] = { '\0' };

int main() {

    /*ofstream fout;
    fout.open("output.txt");*/

    int i,p1,p2,p3,j,l1,l2,l3,tmp;
    seq[0] = seq[1] = 0;
    p1 = 0; p2 = 1; p3 = 2;
    ret[0] = ret[1] = 'a';

    counter_2D[seq[0]][seq[0]]++;
    counter_1D[seq[0]]++;
    counter_1D[seq[1]]++;

    for(i=2;i<1000000;i++) {

        l1 = seq[p1];
        l2 = seq[p2]; 
        for(j=1;j<=26;j++) {
            l3 = (seq[p3] + j) % 26;
            if(counter_3D[l1][l2][l3] < 100 && counter_2D[l2][l3] < 2000 && counter_1D[l3] < 40000) {
                ret[i] = l3 + 'a';
                seq[p3] = l3;
                tmp = p1;
                p1 = p2;
                p2 = p3;
                p3 = tmp;
                counter_3D[l1][l2][l3]++;
                counter_2D[l2][l3]++;
                counter_1D[l3]++;
                break;
            }
        }
    }

    cout<<ret;

    return 0;
}
    /*fout.close();
    ifstream fin;
    fin.open("output.txt",ios_base::in);

    i = 0;
    while(!fin.eof()) 
        fin>>test[i++];

    cnt = strlen(test);
    cout<<"count:"<<cnt<<endl;

    check[test[0]-'a']++;
    check[test[1]-'a']++;
    check2D[test[0] - 'a'][test[1] - 'a']++;

    i = 2;
    while(i < cnt ) {
        check[test[i] - 'a']++;
        check2D[test[i-1] - 'a'][test[i] - 'a']++;
        check3D[test[i-2] - 'a'][test[i-1] - 'a'][test[i] - 'a']++;
        i++;
    }

    for(int i=0;i<26;i++) {
        if(check[i] > 40000) {
            cout<<"1 D exceeds\n"<<endl;
            return 0;
        }
        for(int j=0;j<26;j++) {
            if(check2D[i][j] > 2000) {
                cout<<"2 D exceeds"<<check2D[i][j]<<endl;
                cout<<(char)(i + 'a')<<(char)(j + 'a')<<endl;
                return 0;
            }
            for(int k=0;k<26;k++) {
                if(check3D[i][j][k] > 100) {
                    cout<<"3 D exceeds"<<check3D[i][j][k]<<endl;
                    cout<<(char)(i + 'a')<<(char)(j + 'a')<<(char)(k + 'a')<<endl;
                }
            }
        }
    }

    return 0;
}*/

这题我的基本思路是，每次都根据前面的两个字母，来判断跟第三个字母构成的组合是否能用（超过出现次数就不能了）。

不过这里有个小问题就是，例如aaa, 如果循环下一个找的还是aaa，那么这样很快会耗掉"aaa"这组合的次数，通过验证这样的话构造不了1000000个。

但一个trick就是每次找的时候都从上一个找个字母的后一个字符开始，例如aaa,下一个找aab,然后找abc，bcd这样，马上就OK了 ^_^