课后自主练习（高精度）1082. 猜猜猜 naive《编程思维与实践》个人学习笔记
作者：互联网
题目
思路
显然，长度为1的时候只有一种可能
长度为2的时候，如果1、2单词一致，只有一种可能，1、2单词不一致，那就可以有四种可能
当长度大于等于2的时候
除了s[0]和s[len-1]外，中间的数都得考虑前后的影响
于是我们就特别判断一下首位和末尾
若s[0] == s[1] 可能的情况为1种
s[0]!=s[1]可能的情况为两种
对于末尾的情况同理
于是我们开始讨论s[1]到s[length-1]的情况
倘若是aaa型，那么自然是只有1种符合要求的情况
如果是aab，baa,abat,中间可以变a和b 2种情况（其中aba可能会重复计算，所以只需要判定前两个即可）
如果是abc型，那么中间的满足要求的可以是a，b，c，三种情况
代码(大整数类）
#include<iostream>
using namespace std;
#include<iostream>
#include<cctype>
#include<cstring>
const int MAXN = 3000;
class big
{
public:
int dig[MAXN];
int sign;
int LEN;
//public:
big();
big(string s);
big(const char *c);
big(long long a);
big &operator=(const big& b);
big operator+(big b);
big operator*(big b);
big operator+=(big b);
bool operator==(int a);
void out();
void carry();
void test();
bool com(long long a);
};//
void big::test()
{
cout << sign << " " << LEN << endl;
out();cout << endl;
}
void big::out()
{
int l = MAXN - 1;
while(dig[l] == 0 && l > 0)//注意0的情况
l--;
cout << ((sign == 1) ? "" : "-");
for(;l >= 0; l--)
cout << dig[l];
//cout <<endl;
}
void big::carry()
{
bool flag = 0;
for(int i = 0; i < this->LEN; i++)
{
if(this->dig[i] >= 10)
{
this->dig[i + 1] += this->dig[i] / 10;
this->dig[i] %= 10;
if(i == LEN - 1)
LEN++;
}
else if(this->dig[i] < 0)
{
while(this->dig[i] < 0)
{
this->dig[i + 1]--;
this->dig[i] += 10;
}
flag = 1;
}
}
if(flag)
{
while(dig[LEN - 1] == 0 && LEN - 1 >= 0)
LEN--;
if(LEN == 0)
LEN = 1;
}
}//
big::big()
{
LEN = 1;
sign = 1;
memset(dig,0,sizeof(dig));
}
big::big(string s)
{
memset(dig,0,sizeof(dig));
LEN = 0;
int len = s.length();
for(int i = len - 1; i >= 0; i--)
{
switch(s[i])
{
case '+':
sign = 1;break;
case '-':
sign = -1;break;
default:
dig[LEN++] = s[i] - '0';
break;
}
}
if(sign != -1)sign = 1;
}
big::big(long long a)
{
memset(dig,0,sizeof(dig));
LEN = 0;
sign = ((a >= 0) ? 1 : -1);
if(a < 0)
a = -a;
while(a)
{
dig[LEN++] = a % 10;
a /= 10;
}
if(sign != -1)sign = 1;
}
big::big(const char *c)
{
memset(dig,0,sizeof(dig));
LEN = 0;
int len = strlen(c);
for(int i = len - 1; i >= 0; i--)
{
switch(c[i])
{
case '+':
sign = 1;break;
case '-':
sign = -1;break;
default:
dig[LEN++] = c[i] - '0';
}
}
if(sign != -1)sign = 1;
}
big& big::operator=(const big & b)
{
this->sign = b.sign;
this->LEN = b.LEN;
for(int i = 0; i < LEN;i++)
{
this->dig[i] = b.dig[i];
}
return *this;
}
big big::operator+(big b)
{
if(this->sign * b.sign == -1)
{
if(b.LEN > this->LEN)
{
for(int i = 0; i < b.LEN; i++)
b.dig[i] -= this->dig[i];
b.carry();
return b;
}
else if(b.LEN < this->LEN)
{
for(int i = 0; i < this->LEN; i++)
this->dig[i] -= b.dig[i];
this->carry();
return *this;
}
else
{
bool flag = true;
for(int i = b.LEN - 1; i >= 0; i--)
{
if(this->dig[i] > b.dig[i])
{flag = true;break;}
else if(this->dig[i] < b.dig[i])
{flag = false;break;}
}
if(flag)
{
for(int i = 0; i < this->LEN; i++)
this->dig[i] -= b.dig[i];
this->carry();
return *this;
}
else
{
for(int i = 0; i < b.LEN; i++)
b.dig[i] -= this->dig[i];
b.carry();
return b;
}
}
}
else
{
int mlen = ((this->LEN > b.LEN) ? this->LEN : b.LEN);
for(int i = 0; i < mlen; i++)
this->dig[i] += b.dig[i];
this->carry();
}
return *this;
}
big big::operator*(big b)
{
big B;
B.sign = this->sign * b.sign;
for(int i = 0; i < b.LEN; i++)
{
for(int j = 0; j < this->LEN; j++)
{
B.dig[i + j] += this->dig[j] * b.dig[i];
B.LEN = i + j + 1;//注意长度
B.carry();
}
}
return B;
}
bool big::operator==(int a)
{
if(a * sign < 0)
return false;
if(a < 0)
a = -a;
int alen = 0;
int dlen = MAXN;
while(dig[dlen - 1] == 0 && dlen - 1 >= 0)
dlen--;
if(dlen == 0)
dlen++;
int temp = a;
if(temp == 0)
alen = 1;
while(temp)
{
alen++;
temp/=10;
}
if(dlen != alen)
return false;
temp = 0;
for(int i = 0; i < alen; i++)
{
if(a % 10 != dig[i])
return false;
a/=10;
temp++;
}
return true;
}
void out(big R, big I)
{
if(R==0)
{
if(I==0)
cout << "0";
else if(I==1)
cout << "i";
else if(I==-1)
cout << "-i";
else
{
I.out();
cout << 'i';
}
}
else
{
R.out();
if(I==0){}
else if(I==1)
cout << "+i";
else if(I==-1)
cout << "-i";
else
{
cout << ((I.sign == 1) ? "+" : "");
I.out();
cout << 'i';
}
}
}
代码（主体）
const int MOD = 1e9+7;
typedef long long ll;
big func(string a)
{
int len = a.length();
if(len == 1) return 1;
if(len == 2) return ((a[0] == a[1]) ? 1 : 4);
big r{1};
if(a[0] != a[1]) r = 2;
for(int i = 1; i < len - 1; i++)
{
big c{1};
if(a[i] != a[i - 1]) c=c+1;
if(a[i] != a[i + 1] && a[i - 1] != a[i + 1]) c=c+1;
r = (r * c) ;
}
if(a[len - 1] != a[len - 2]) r = r * 2;
return r;
}
//
int main()
{
string s;
cin >> s;
func(s).out();
}
标签：1082,猜猜猜,int,big,dig,LEN,sign,++,课后 来源： https://blog.csdn.net/JamSlade/article/details/116405613