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Educational Codeforces Round 147 (Rated for Div. 2)

发布时间 2023-04-21 22:22:16作者: harper886

Educational Codeforces Round 147 (Rated for Div. 2)

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Educational Codeforces Round 147 (Rated for Div. 2)

A题

  1. 如果第一位数是0,直接打印0

  2. 如果第一位数是'?',有9个数可以选择,如果其他位数是'?',有10中情况选择,相乘就可以了

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cstring>
#include <unordered_set>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <sstream>
#include <queue>
#define int long long
#define yes cout<<"YES"<<'\n'
#define no 	cout<<"NO"<<'\n'

using namespace std;
const int N = 100008;

void solve() {

	string s;
	cin >> s;
	if (s[0] == '0') {//第1位是0直接打印0
		cout << 0 << '\n';
		return;
	}
	int sum = 1;
	for (int i = 0; i < (int)s.size(); i++) {
		if (s[i] == '?') {//如果第一位是?,有9中情况可以选择
			if (i == 0) {
				sum = sum * 9;
			} else {
				sum = sum * 10;//其他如果是?,有10中情况选择
			}
		}
	}
	cout << sum << '\n';//打印






}
signed main () {
	std::ios::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);
	int t;
	cin >> t;
	while (t) {
		solve();
		t--;
	}


	return 0;
}

B题

  1. 先找到两个数组第一个不同的数为l,然后找到两个数组最后一个不同的数为r.
  2. 然后判断l,和r能不能继续扩展,如果可以把l 和 r都扩展到最大
  3. 打印l和r
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cstring>
#include <unordered_set>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <sstream>
#include <queue>
#define int long long
#define yes cout<<"YES"<<'\n'
#define no 	cout<<"NO"<<'\n'

using namespace std;
const int N = 200008;
int a[N];
int b[N];
void solve() {
//	int ans = 0;
	int n;
	scanf("%lld", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);

	}
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &b[i]);
	}
	int l = 0, r = 0;
	for (int i = 1; i <= n; i++) {
		if (a[i] != b[i]) {
			l = i;//找到l
			break;
		}
	}
	for (int i = n; i >= 1; i--) {
		if (a[i] != b[i]) {
			r = i;//找到r
			break;
		}
	}
	int mmin = min(a[l], b[l]);//取a[l]和b[l]的较小值
	int mmax = max(a[r], b[r]);//取a[l]和b[l]的较大值
	a[l] = mmin;
	b[l] = mmin;//对两个数组赋值
	for (int i = l; i >= 2; i--) {//如果前一个数小于当前的数,就可以向前扩展
		if (a[i] >= a[i - 1]) {
			l--;
		} else {
			break;//不满足直接跳出
		}
	}
	a[r] = mmax;
	b[r] = mmax;
	for (int i = r; i <= n - 1; i++) {//如果后一个数大于当前的数,就可以向后扩展
		if (a[i] <= a[i + 1]) {
			r++;
		} else {
			break;//不满足直接跳出
		}
	}

	cout << l << ' ' << r << '\n';//打印l和r




}
signed main () {
//	std::ios::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);
	int t;
	cin >> t;
	while (t) {
		solve();
		t--;
	}


	return 0;
}