2020 Multi-University Training Contest 10（待补
2020 Multi-University Training Contest 10
1003 Mine Sweeper
-
思路：随机数生成
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int d[8][2] = {1, 0, 0, 1, -1, 0, 0, -1, -1, -1, -1, 1, 1, -1, 1, 1};
int t, s;
map<int, vector<vector<char> >> mp;
void init(){
srand(time(NULL));
for (int k = 1; k <= 50000; k ++ ){
int r = (rand() % 25) + 1, c = (rand() % 25) + 1;
vector<vector<char> > ans(r, vector<char>(c));
for (int i = 0; i < r; i ++ )
for (int j = 0; j < c; j ++ )
ans[i][j] = (rand() % 2) ? 'X' : '.';
int sum = 0;
for (int i = 0; i < r; i ++ ){
for (int j = 0; j < c; j ++ ){
if (ans[i][j] == '.'){
int cnt = 0;
for (int l = 0; l < 8; l ++ ){
int dx = i + d[l][0], dy = j + d[l][1];
if ((dx >= 0 && dx <= r - 1) && (dy >= 0 && dy <= c - 1) && ans[dx][dy] == 'X')
cnt ++ ;
}
sum += cnt;
}
}
}
mp[sum] = ans;
}
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
init();
cin >> t;
while (t -- ){
cin >> s;
vector<vector<char> > ans = mp[s];
int r = ans.size(), c = ans[0].size();
cout << r << " " << c << "\n";
for (int i = 0; i < r; i ++ ){
for (int j = 0; j < c; j ++ )
cout << ans[i][j];
cout << "\n";
}
}
return 0;
}
1004 Permutation Counting
- 思路：
\[b = 0 \\
dp[i][j] = \sum_{k=1}^{j-1}dp[i-1][k] \\
b = 1 \\
dp[i][j] = \sum_{k=j+1}^{k=i}dp[i-1][k] \\
\]
再将复杂度降成\(O(n^2)即可\)
- AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 5010;
const int mod = 1e9 + 7;
int t, n, b, ans;
ll dp[N][N];
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
ans = 0;
cin >> n;
for (int i = 1; i < n; i ++ ){
cin >> b;
if (b == 0)
for (int j = 1; j <= i; j ++ )
dp[i][j] = (dp[i][j - 1] + dp[i - 1][j - 1]) % mod;
else
for (int j = i - 1; j >= 0; j -- )
dp[i][j] = (dp[i][j + 1] + dp[i - 1][j]) % mod;
}
for (int i = 0; i < n; i ++ )
ans = (1ll * ans + dp[n - 1][i]) % mod;
cout << ans << "\n";
}
return 0;
}
1011 Task Scheduler
-
思路：若\(k\)不为0，按t降序排列后输出原次序，若 \(k\)为0，输出\(1-n\)
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 110;
int T, n, m, k;
struct node{
int t, id;
}a[N];
inline bool cmp(node p, node q){
if (p.t != q.t)
return p.t > q.t;
return p.id < q.id;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> T;
while (T -- ){
cin >> n >> m >> k;
for (int i = 1; i <= n; i ++ ){
cin >> a[i].t;
a[i].id = i;
}
if (k)
sort(a + 1, a + n + 1, cmp);
for (int i = 1; i < n; i ++ )
cout << a[i].id << " ";
cout << a[n].id << "\n";
}
return 0;
}
