一:
直接输出, 用flag记录是否有答案,一次循环,时间复杂度O(n)。
代码
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
string s;
int main()
{
while(cin >> s)
{
if(s.size() == 1)
{
cout << s << endl;
continue;
}
bool flag = false;
for(int i = 0; i < s.size() - 1; i++)
{
if(s[i] != s[i + 1])
{
cout << s[i];
flag = true;
}
else
{
char t = s[i];
while(s[i] == t)
i++;
i--;
}
}
if(!flag)
cout <<"no" << endl;
else
{
if(s[s.size() - 1] != s[s.size() - 2])
cout << s[s.size() - 1] <<endl;
else cout << endl;
}
s = "";
}
return 0;
}
二:
字符串枚举。
代码
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
string s1, s2;
string getMinString(string str1, string str2)
{
string ans;
if(str2.size() == 0)
return str1;
for(int i = 0; i < str1.size(); i++)
{
int a[110];
for(int i = 0; i <= str2.size(); i++)
a[i] = 1;
int tol = str2.size();
for(int j = i; j >= 0; j--)
{
for(int k = 0; k < str2.size(); k++)
{
if(str2[k] == str1[j] && a[k] == 1)
{
a[k]--;
tol--;
break;
}
}
if(tol == 0)
{
string t = str1.substr(j, i - j + 1);
//cout << t << endl;
if(t.size() < ans.size() || ans.size() == 0)
ans = t;
}
}
}
return ans;
}
int main()
{
cin >> s1 >> s2;
cout << getMinString(s1, s2) << endl;
return 0;
}
三:
交叉输出就行
代码
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1000010;
int b[N], g[N];
int n;
int main()
{
cin >> n;
for(int i = 0; i < n; i++)
cin >> b[i];
for(int i = 0; i < n; i++)
cin >> g[i];
for(int i = 0; i < n; i++)
{
cout << b[i] << endl;
cout << g[i] << endl;
}
return 0;
}
四:
暴力枚举
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1000010;
int ax[30][30];
int main()
{
char c;
int col = 0, row = 0;
int j;
char last;
int k = 1;
bool change = false;
while(cin >> c)
{
//cout << c <<endl;
if(k == 1)
{
k++;
continue;
}
if(c == '}' && last == '}')
break;
if(c == '{')
{
row++;
j = 1;
}
else if(c == '1')
{
ax[row][j] = 1;
col = max(col, j);
j++;
}
else if(c == '0')
{
ax[row][j] = 0;
col = max(col, j);
j++;
}
last = c;
}
int ans = 0;
for(int a = 1; a <= row; a++)
for(int b = 1; b <= col; b++)
for(int c = 1; c <= a; c++)
for(int d = 1; d <= b; d++)
{
int t = 0;
bool zore = true;
for(int e = c; e <= a && zore; e++)
for(int g = d; g <= b && zore; g++)
if(ax[e][g] == 1 && zore)
t++;
else
zore = false;
if(zore)
ans = max(ans, t);
}
cout << ans << endl;
return 0;
}