2020 Multi-University Training Contest 7（待补
2020 Multi-University Training Contest 7
1007 Game
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思路：依次取最长边，最后只剩下起点一个点，后手可以选择先手选的所在边的另一个点，使先手败
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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 = 2010;
int t, n, tot;
bool vis[N];
ll x[N], y[N];
struct node{
int u, v;
ll dis;
}a[N * N];
inline bool cmp(node p, node q){
return p.dis < q.dis;
}
ll dist(int i, int j){
return (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
}
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 -- ){
tot = 0;
memset(vis, false, sizeof(vis));
cin >> n;
for (int i = 1; i <= n; i ++ )
cin >> x[i] >> y[i];
for (int i = 1; i <= n; i ++ ){
for (int j = i + 1; j <= n; j ++ ){
a[ ++ tot ].u = i;
a[tot].v = j;
a[tot].dis = dist(i, j);
}
}
sort(a + 1, a + tot + 1, cmp);
for (int i = tot; i >= 1; i -- ){
if (!vis[a[i].u] && !vis[a[i].v]){
vis[a[i].u] = true;
vis[a[i].v] = true;
}
}
if (vis[1])
cout << "YES\n";
else
cout << "NO\n";
}
return 0;
}
1009 Increasing and Decreasing
-
思路：贪心
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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 = 1e5 + 10;
int t, pos, tot;
int ans[N];
ll n, x, y;
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 -- ){
pos = tot = 0;
cin >> n >> x >> y;
if (x * y + 1 <= n || x + y - 1 > n){
cout << "NO\n";
continue;
}
cout << "YES\n";
for (int i = 1; i <= n; i ++ ){
if ((x - 1) * y + i < n)
continue;
for (int j = i; j > pos; j -- )
ans[ ++ tot ] = j;
pos = i;
x -- ;
}
for (int i = 1; i < n; i ++ )
cout << ans[i] << " ";
cout << ans[n] << "\n";
}
return 0;
}
1010 Jogging
-
思路：暴力bfs出所有点建无向图，答案就是“起点的度数+1”/“总度数+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;
}
typedef pair<ll, ll> pll;
int t, ans1, ans2;
ll x, y;
map<pll, int> vis;
inline bool bfs(ll x, ll y){
queue<pll> q;
q.push(make_pair(x, y));
vis[make_pair(x, y)] = 1;
while (!q.empty()){
pll tmp = q.front();
q.pop();
if (x == y)
return false;
ans2 ++ ;
for (int i = -1; i <= 1; i ++ ){
for (int j = -1; j <= 1; j ++ ){
if (i == 0 && j == 0)
continue;
ll dx = tmp.first + i, dy = tmp.second + j;
if (dx == dy)
return false;
if (gcd(dx, dy) != 1){
ans2 ++ ;
if (!vis[make_pair(dx, dy)]){
vis[make_pair(dx, dy)] = 1;
q.push(make_pair(dx, dy));
}
}
}
}
if (!ans1)
ans1 = ans2;
}
return true;
}
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 -- ){
ans1 = ans2 = 0;
vis.clear();
cin >> x >> y;
if (!bfs(x, y))
cout << "0/1\n";
else
cout << ans1 / gcd(ans1, ans2) << "/" << ans2 / gcd(ans1, ans2) << "\n";
}
return 0;
}
