洛谷训练4：字符串训练

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输入 #1复制
10
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输出 #1复制
5

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#include<iostream>
#include<vector>
using namespace std;

int f(vector<int>& vec) {
	int n = vec.size();
	int* dp = new int[n];
	for (int i = 0; i < n; i++) {
		dp[i] = 0;
	}
	dp[0] = 1;
	int prev,sum;
	for (int j = 1; j < n; j++) {
		prev = dp[j - 1];
		sum = 0;
		for (int i = j; i > 0; i--) {
			sum++;
			if (vec[i] < vec[i - 1]) {
				break;
			}
		}
		dp[j] = prev > sum ? prev : sum;
	}
	return dp[n-1];
}

int f2(vector<int>& vec) {
	int n = vec.size();
	int* dp = new int[n];
	int k = 1;
	dp[0] = 1;
	for (int i = 1; i < n; i++) {
		if (vec[i] > vec[i - 1]) {
			k++;
			dp[i] = k;
		}
		else {
			k = 1;
			dp[i] = k;
		}
	}
	int max = 1;
	for (int i = 0; i < n; i++) {
		max = max > dp[i] ? max : dp[i];
	}
	return max;
}

int f3(int a[],int N) {
	int n = N;
	int* dp = new int[n];
	int k = 1;
	dp[0] = 1;
	for (int i = 1; i < n; i++) {
		if (a[i] > a[i - 1]) {
			k++;
			dp[i] = k;
		}
		else {
			k = 1;
			dp[i] = k;
		}
	}
	int max = 1;
	for (int i = 0; i < n; i++) {
		max = max > dp[i] ? max : dp[i];
	}
	return max;
}

int f4(int a[], int N) {
	int n = N;
	int* dp = new int[n];
	for (int i = 0; i < n; i++) {
		dp[i] = 0;
	}
	dp[0] = 1;
	int prev, sum;
	for (int j = 1; j < n; j++) {
		prev = dp[j - 1];
		sum = 0;
		for (int i = j; i > 0; i--) {
			sum++;
			if (a[i] < a[i - 1]) {
				break;
			}
		}
		dp[j] = prev > sum ? prev : sum;
	}
	return dp[n - 1];
}

int main() {
	int N;
	int num;
	cin >> N;
	/*
	vector<int> vec;
	for (int i = 0; i < N; i++) {
		cin >> num;
		vec.push_back(num);
	}
	*/
	int* arr = new int[N];
	for (int i = 0; i < N; i++) { cin >> arr[i]; }
	int count = f3(arr,N);
	cout << count;
	return 0;
}

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输入 #1复制
0-670-82162-4
输出 #1复制
Right
输入 #2复制
0-670-82162-0
输出 #2复制
0-670-82162-4

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#include<iostream>
using namespace std;
char num[14];

int main() {
	char mod[] = "0123456789X";
	cin >> num;
	int val = 0;
	int mult = 1;
	for (int i = 0; i < 11; i++) {
		if (num[i] == '-')continue;
		val += (num[i] - '0')*mult;
		mult++;
	}
	val = val % 11;
	char rightans = mod[val];
	if (rightans == num[12])cout << "Right";
	else {
		num[12] = rightans;
		cout << num;
	}
	return 0;
}

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#include<iostream>
using namespace std;

int f(char a[]) {
	int res = 1;
	for (int i = 0; a[i] != ''; i++) {
		res *= (a[i] - 'A'+1);
	}
	res = res % 47;
	return res;
}

int main() {
	char name1[7];
	char name2[7];
	cin >> name1;
	cin >> name2;
	int n1, n2;
	n1 = f(name1);
	n2 = f(name2);
	n1 == n2 ? cout << "GO" : cout << "STAY";
	return 0;
}

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#include<iostream>
#include<string>
#include<algorithm>
using namespace std;

string reverse2(string s) {
	/*将一段字符串翻转并去除前面的0*/
	reverse(s.begin(), s.end());
	int zerocount = 0;
	for (auto i : s) {
		if (i == '0')zerocount++;
		else break;
	}
	s.erase(s.begin(), s.begin() + zerocount);
	return (s != "" ? s:"0");
}

string ReverseTail(string s) {
	/*将一段字符串翻转并去掉末尾的0*/
	reverse(s.begin(), s.end());
	int zerocount = 0;
	for (auto iter = s.end() - 1; iter != s.begin(); iter--) {
		if (*iter == '0')zerocount++;
		else break;
	}
	s.erase(s.end() - zerocount, s.end());
	return (s != "" ? s : "0");
}


int main() {
	string s;
	cin >> s;
	if (s.back() == '%') {
		cout<<reverse2(s.substr(0,s.size()-1));
		return 0;
	}
	if (s.find('.')!=std::string::npos ) {
		int pos = s.find('.');
		string left = reverse2(s.substr(0, pos));
		string right = ReverseTail(s.substr(pos + 1));
		cout << left << "." << right;
		return 0;
	}

	if (s.find('/') != std::string::npos) {
		int pos = s.find('/');
		string left=reverse2(s.substr(0, pos));
		string right= ReverseTail(s.substr(pos + 1));
		cout << left << "/" << right;
		return 0;
	}

	cout << reverse2(s);
	return 0;
}

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#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int main() {
	int n;
	char input[51];
	cin >> n;
	cin >> input;
	int N = 'z' - 'a' + 1;
	n = n%N;
	int len = strlen(input);
	for (int i = 0; i < len; i++) {
		if (input[i]+n>'z')input[i] =input[i]+n- 'z'+'a'-1;
		else input[i]+=n;
	}
	cout << input;
	return 0;
}