#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 ^_^
