8.18 2020 Multi-University Training Contest 9题解及补题
目录
8.18 2020 Multi-University Training Contest 9题解及补题
比赛过程
据说最后两场比较难，改了很多次出的A，就相当于费劲签了到。
题解
A Tree
题意
给了你一棵树，然后加一条有向边，找到最大数量的\((x,y)\)。
解法
假如说没加边，那么总共的数量就是所有节点的孩子数量(包括自己)全部加起来，加入的这条有向边一定是从某一叶子节点到根节点，接着开个数组存一下信息就可以开始乱搞了。
比赛的时候一直TLE和WA
代码
#pragma region
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, n) for (int i = a; i <= n; ++i)
#define per(i, a, n) for (int i = n; i >= a; --i)
namespace fastIO
{
#define BUF_SIZE 100000
#define OUT_SIZE 100000
//fread->R
bool IOerror = 0;
//inline char nc(){char ch=getchar();if(ch==-1)IOerror=1;return ch;}
inline char nc()
{
static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
if (p1 == pend)
{
p1 = buf;
pend = buf + fread(buf, 1, BUF_SIZE, stdin);
if (pend == p1)
{
IOerror = 1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) { return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t'; }
template <class T>
inline bool R(T &x)
{
bool sign = 0;
char ch = nc();
x = 0;
for (; blank(ch); ch = nc())
;
if (IOerror)
return false;
if (ch == '-')
sign = 1, ch = nc();
for (; ch >= '0' && ch <= '9'; ch = nc())
x = x * 10 + ch - '0';
if (sign)
x = -x;
return true;
}
inline bool R(double &x)
{
bool sign = 0;
char ch = nc();
x = 0;
for (; blank(ch); ch = nc())
;
if (IOerror)
return false;
if (ch == '-')
sign = 1, ch = nc();
for (; ch >= '0' && ch <= '9'; ch = nc())
x = x * 10 + ch - '0';
if (ch == '.')
{
double tmp = 1;
ch = nc();
for (; ch >= '0' && ch <= '9'; ch = nc())
tmp /= 10.0, x += tmp * (ch - '0');
}
if (sign)
x = -x;
return true;
}
inline bool R(char *s)
{
char ch = nc();
for (; blank(ch); ch = nc())
;
if (IOerror)
return false;
for (; !blank(ch) && !IOerror; ch = nc())
*s++ = ch;
*s = 0;
return true;
}
inline bool R(char &c)
{
c = nc();
if (IOerror)
{
c = -1;
return false;
}
return true;
}
template <class T, class... U>
bool R(T &h, U &... t) { return R(h) && R(t...); }
#undef OUT_SIZE
#undef BUF_SIZE
}; // namespace fastIO
using namespace fastIO;
template <class T>
void _W(const T &x) { cout << x; }
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) { _W(x.fa), putchar(' '), _W(x.S); }
template <class T>
void _W(const vector<T> &x)
{
for (auto i = x.begin(); i != x.end(); _W(*i++))
if (i != x.cbegin())
putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &... tail) { _W(head), putchar(sizeof...(tail) ? ' ' : '\n'), W(tail...); }
#pragma endregion
const int maxn = 5e5 + 500;
const int mod = 1e9 + 7;
ll gcd(ll T, ll b) { return b == 0 ? T : gcd(b, T % b); }
ll lcm(ll T, ll b) { return T / gcd(T, b) * b; }
ll mul(ll a, ll b, ll c)
{
ll ans = 0;
while (b)
{
if (b & 1)
{
ans = (ans + a) % c;
b--;
}
b >>= 1;
a = (a + a) % c;
}
return ans;
}
ll powmod(ll a, ll b, ll c)
{
ll ans = 1;
while (b)
{
if (b & 1)
{
ans = mul(ans, a, c);
b--;
}
b >>= 1;
a = mul(a, a, c);
}
return ans;
}
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
ll fa[maxn];
vector<ll> G[maxn];
ll num[maxn], sum[maxn];
ll n, ans = 0;
void dfs(ll u, ll f)
{
++num[u];
for (auto v : G[u])
{
dfs(v, u);
num[u] += num[v];
}
ans += num[u];
}
void dfs2(ll u, ll f)
{
sum[u] = sum[f] + n - num[u];
for (auto v : G[u])
{
dfs2(v, u);
}
}
void init()
{
fa[1] = 0;
for (int i = 1; i <= n; ++i)
{
sum[i] = 0, num[i] = 0, G[i].clear();
}
}
int main()
{
ll T;
scanf("%lld", &T);
while (T--)
{
ans = 0;
scanf("%lld", &n);
init();
for (int i = 2; i <= n; ++i)
{
scanf("%lld", fa + i);
G[fa[i]].push_back(i);
}
dfs(1, 0);
dfs2(1, 0);
sort(sum + 1, sum + 1 + n);
ans += sum[n];
printf("%lld\n", ans);
}
return 0;
}
B
题意
解法
代码
//将内容替换成代码
C Slime and Stones
题意
有两堆石子，第一堆有 $a \(个石头，第二堆有\) b$个石头。
你有两种操作：
操作一：从某堆拿任意个石子。
操作二：从第一堆拿 $x $个，从第二堆拿 \(y\)个，要求 \(∣x−y∣≤k\),其中\(k\)是给定常数。
每个人都是最优策略，让你求是否先手必胜（1）或先手必败（2）
解法
代码
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#define IO ios::sync_with_stdio(0), cin.tie(0)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf = ~0u >> 1;
typedef pair<int, int> P;
#define REP(i, a, n) for (int i = a; i < (n); ++i)
#define PER(i, a, n) for (int i = (n)-1; i >= a; --i)
const ll mod = 1e9 + 7;
ll gcd(ll T, ll b) { return b == 0 ? T : gcd(b, T % b); }
ll lcm(ll T, ll b) { return T / gcd(T, b) * b; }
const int maxn = 1000002;
ll a[maxn];
void init() {
a[0]=1;
for(int i=1;i<=1000000;i++) {
a[i]=(a[i-1]*i)%mod;
}
}
template <typename T>
void read(T &x)
{
x = 0;
char ch = getchar();
ll f = 1;
while (!isdigit(ch))
{
if (ch == '-')
f *= -1;
ch = getchar();
}
while (isdigit(ch))
{
x = x * 10 + ch - 48;
ch = getchar();
}
x *= f;
}
int main()
{
int T;
cin>>T;
while(T--){
int a,b,t;
cin>>a>>b>>t;
double k = (double)t;
if(a > b)swap(a,b);
double XA = (1.0-k+sqrt(5.0+2*k+k*k))/2;
double XB = (3.0+k+sqrt(5.0+2*k+k*k))/2;
int l = 0;
int r = 100000000;
while(l < r){
int mid = l + r >> 1;
if(floor((double)mid * XA) > a)r = mid - 1;
else if(floor((double)mid * XA) < a)l = mid + 1;
else {
l = mid;
break;
}
}
if(floor((double)l * XA) == a && floor((double)l * XB) == b)printf("0\n");
else printf("1\n");
}
return 0;
}
D
题意
解法
代码
//将内容替换成代码
E
题意
解法
代码
//将内容替换成代码
F
题意
解法
代码
//将内容替换成代码
G Game
题意
解法
代码
//将内容替换成代码
H
题意
解法
代码
//将内容替换成代码
I
题意
解法
代码
//将内容替换成代码
J
题意
解法
代码
//将内容替换成代码
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