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;elsesum += 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;
}
