目录
1.字符串替换

简单的遍历替换即可:
cpp
class Solution {
public:
string formatString(string str, vector<char>& arg) {
string ret;
int k = 0;
for (int i = 0; i < str.size(); ++i)
{
if (str[i] == '%')
{
ret += arg[k++];
++i;
}
else
{
ret += str[i];
}
}
if (k != arg.size())
{
while (k != arg.size())
{
ret += arg[k++];
}
}
return ret;
}
};
替换完之后别忘了后面的剩余字符。
2.神奇数

一个检查是否为质数的函数,一个检查是否为神奇数的函数。
然后for循环遍历数组即可。
cpp
#include <iostream>
using namespace std;
bool Check_zs(int x) {
if (x == 1)
return false;
for (int i = 2; i < x; ++i)
if (x % i == 0)
return false;
return true;
}
bool Check_sq(int x) {
int arr[10] = { 0 };
int cnt = 0;
int k = 0;
while (x) {
arr[k++] = x % 10;
x /= 10;
cnt++;
}
for (int i = 0; i < cnt; ++i) {
for (int j = i + 1; j < cnt; ++j) {
if (Check_zs(arr[i] * 10 + arr[j]) && arr[i] != 0) {
return true;
}
}
}
for (int i = cnt - 1; i >= 0; --i) {
for (int j = i - 1; j >= 0; --j) {
if (Check_zs(arr[i] * 10 + arr[j]) && arr[i] != 0) {
return true;
}
}
}
return false;
}
int main() {
int l, r;
cin >> l >> r;
int cnt = 0;
for (int i = l; i <= r; ++i) {
if (Check_sq(i)) {
cnt++;
}
}
cout << cnt << endl;
return 0;
}
注意细节:不要出现前导0。

3.DNA序列

cpp
#include <iostream>
#include <vector>
using namespace std;
vector<pair<int, float>> vp(1010);
int main() {
string s;
int n;
cin >> s >> n;
int l = 0;
int k = 0;
while (l != s.size() - n + 1)
{
int cnt = 0;
for (int i = l; i < l + n; ++i)
if (s[i] == 'C' || s[i] == 'G')
cnt++;
vp[k++] = { l, (float)cnt / n };
l++;
}
int keyi = 0;
float Max = 0.0;
for (int i = 0; i < k; ++i)
if (vp[i].second > Max)
{
keyi = vp[i].first;
Max = vp[i].second;
}
for (int i = keyi; i < keyi + n; ++i)
cout << s[i];
cout << endl;
return 0;
}
我的思路是遍历所有长度为n的子串,找到比例最高的一串。
类似固定长度滑动窗口。