A. X Axis

 

思路:

1~10暴力枚举一下所有可能

代码:

#include<bits/stdc++.h>
using namespace std;
#define N 1000005
typedef long long ll;
typedef unsigned long long ull;
ll n, m, t, h, k;
ll a, b, c;
ll ans, num, sum, cnt;
ll temp[N], f1[N], f2[N];
bool flag, vis[N];
int main()
{
	cin >> t;
	while (t--) {
		cin >> a >> b >> c;
		sum = a + b + c;
		ans = 1e9;
		for (int i = 0; i <= 10; i++) {
			ans = min(ans,abs(a-i)+abs(b-i)+abs(c-i));
		}
		cout << ans << endl;
	}
}

B. Matrix Stabilization

思路:

若一个数比周围四个数都大，那么最终会变成四个数当中最大的哪一个,我们只要将四周的数的最大值找到,然后把这个数变成相邻的最大值.

代码:

#include<bits/stdc++.h>
using namespace std;
#define N 1000005
typedef long long ll;
typedef unsigned long long ull;
ll n, m, t, h, k;
ll a, b, c;
ll ans, num, sum1,sum,sum2, cnt;
ll temp[N], f1[N], f2[N];
ll mp[105][105];
bool flag, vis[N];
int main()
{
	cin >> t;
	while (t--) {
		cin >> n >> m;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				cin >> mp[i][j];
			}
		}
		for (int i = 1; i <= n; i++) {
			mp[i][0] = 0, mp[i][m + 1] = 0;
		}
		for (int j = 1; j <= m; j++) {
			mp[0][j] = 0, mp[n + 1][j] = 0;
		}
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				sum1 = max(mp[i - 1][j], mp[i + 1][j]);
				sum2 = max(mp[i][j + 1], mp[i][j - 1]);
				sum = max(sum1, sum2);
				if (mp[i][j] > sum) {
					mp[i][j] = sum;
				}
			}
		}
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				cout << mp[i][j] << " ";
			}
			cout << endl;
		}
	}
	return 0;
}

C. Update Queries

 

思路:

对字符串c以及ind数组进行排序，通过贪心可以知道，我们需要按照索引从小到大的修改字符串，同时一个位置只会有一个字母与之对应，因此只需要同时按照字符串c从小到大修改即可.

代码:

#include<bits/stdc++.h>
using namespace std;
#define N 1000005
typedef long long ll;
typedef unsigned long long ull;
ll n, m, t, h, k;
ll a, b, c;
ll ans, num, sum1,sum,sum2, cnt;
ll temp[N], f1[N], f2[N];
ll mp[105][105];
bool flag, vis[N];
string s, ss;
int main()
{
	cin >> t;
	while (t--) {
		map<ll, ll>mp;
		cin >> n >> m;
		cin >> s;
		for (int i = 1; i <= m; i++) {
			cin >> temp[i];
			mp[temp[i]] = 1;
		}
		cin >> ss;
		s = " " + s;
		ss = " " + ss;
		sort(ss.begin(), ss.end());
		//cout << ss << endl;
		num = 0;
		for (int i = 1; i <= n; i++) {
			if (mp[i]) {
				s[i] = ss[++num];
			}
		}
		for (int i = 1; i <= n; i++)
			cout << s[i];
		cout << endl;
	}
	return 0;
}

D. Mathematical Problem

 

思路:

一定有一个两位数，因此我们可以枚举这个两位数，然后取最小值。接下来考虑如何取最小值：若存在数字0，那么可以通过都用乘法来将最终结果变成0，此外，若存在数字1，可以通过乘法将1消掉。除此之外的数，都是加法更加小。

代码:

#include<bits/stdc++.h>
using namespace std;
#define N 1000005
typedef long long ll;
typedef unsigned long long ull;
ll n, m, t, h, k;
ll a, b, c;
ll ans, num, sum1,sum,sum2, cnt;
ll temp[N], f1[N], f2[N];
ll mp[105][105];
bool flag, vis[N];
string s, ss;
void solve()
{
	for (int i = 1; i < n; i++) {
		vector<ll>v;
		sum = 0;
		for (int j = 0; j < n; j++) {
			if (j + 1 == i) {
				v.push_back(temp[j] * 10 + temp[j + 1]);
				j++;
			}
			else {
				v.push_back(temp[j]);
			}
		}
		for (auto kk : v) {
			if (kk == 0) {
				cout << 0 << endl;
				return;
			}
			if (kk == 1)
				continue;
			else
				sum += kk;
		}
		if (sum == 0)
			sum++;
		ans = min(ans, sum);
	}
	cout << ans << endl;
}
int main()
{
	cin >> t;
	while (t--) {
		cin >> n;
		cin >> s;
		ans = 1e9;
		for (int i = 0; i < n; i++)
			temp[i] = s[i] - '0';
		solve();
	}
	return 0;
}