解题思路:对字符串进行翻转后,求一下最长公共子序列就是最长的回文串,用总长度减去最长回文串的长度即为要删去的字符数
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
int dp[1005][1005];
int LCS(char str[], char re[], int l)
{
for (int i = 0 ; i <= l ; i++)
dp[i][0] = dp[0][i] = 0;
for (int i = 1 ; i <= l ; i++)
{
for (int j = 1 ; j <= l ; j++)
{
if (str[i-1] == re[j-1])
dp[i][j] = dp[i-1][j-1] + 1;
else
dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
}
}
return dp[l][l];
}
int main()
{
char str[1005];
char re[1005];
int l;
while(~scanf("%s",str))
{
l = strlen(str);
for (int i=0, j=l-1 ; i < l ; i++, j--)
re[i] = str[j];
re[l] = '\0';
printf("%d\n",l-LCS(str,re,l));
}
return 0;
}
解题思路:直接扫描一遍,把小写字母挨着往前移即可。
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
bool isUpper(char c)
{
if (c >= 'A' && c <= 'Z')
return true;
return false;
}
int main()
{
char str[1005];
while(~scanf("%s",str))
{
int l = strlen(str);
int cnt = 0; //大写字母数
for (int i = 0 ; i < l ; i++)
{
if (isUpper(str[i]))
cnt++;
else
{
char temp = str[i];
for (int j=0, pos=i ; j < cnt ; j++, pos--)
str[pos] = str[pos-1];
str[i-cnt] = temp;
}
}
printf("%s\n",str);
}
return 0;
}
解题思路:这个要考虑的情况多一点,对于第一个结果,要考虑是否有相同的数字;对于第二个结果,要考虑是否所有数字都一样。
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x7fffffff
#define LL long long
int n;
int num[100000+5];
int main()
{
while(~scanf("%d",&n))
{
for (int i = 0 ; i < n ; i++)
scanf("%d",&num[i]);
sort(num, num+n);
pair<int, int> ans1(INF, 0);
pair<int, int> min_num(INF, 0);
pair<int, int> max_num(-1, 0);
for (int i = 0 ; i < n ; i++)
{
//统计最小差
if (i != 0)
{
if (num[i]-num[i-1] == ans1.first)
ans1.second++;
else if (num[i]-num[i-1] < ans1.first)
{
ans1.first = num[i]-num[i-1];
ans1.second = 1;
}
}
//统计最大最小值
if (num[i] == min_num.first)
min_num.second++;
else if (num[i] < min_num.first)
{
min_num.first = num[i];
min_num.second = 1;
}
if (num[i] == max_num.first)
max_num.second++;
else if (num[i] > max_num.first)
{
max_num.first = num[i];
max_num.second = 1;
}
}
LL p1, p2;
p1 = p2 = 0;
if (ans1.first == 0)
{
ans1.first = num[0];
ans1.second = 1;
for (int i = 1 ; i < n ; i++)
{
if (num[i] == ans1.first)
ans1.second++;
else
{
p1 += (LL)ans1.second*(ans1.second-1)/2;
ans1.first = num[i];
ans1.second = 1;
}
}
p1 += (LL)ans1.second*(ans1.second-1)/2;
}
else
p1 = ans1.second;
if (max_num.first == min_num.first)
p2 = (LL)min_num.second*(min_num.second-1);
else
p2 = (LL)min_num.second*max_num.second;
printf("%lld %lld\n",p1,p2);
}
return 0;
}