ABC201题解

因为学校的某些操作。
让最近两天的(Atcoder)都能打了，挺开心的。
想上次(ABC)看错题意，失败退场。
——————————————————————————————
(A)
直接手动判断六种排列。

A

#include<iostream>
#include<cstdio>
#define ll long long 

ll a,b,c;

int main(){
  scanf("%lld%lld%lld",&a,&b,&c);
  //abc acb bac bca cab cba
  if((b - a) == (c - b) || (c - a) == (b - c) || (a - b) == (c - a) || (c - b) == (a - c) || (a - c) == (b - a) || (b - c) == (a - b))
  puts("Yes");
  else
  puts("No");
}

(B)

B

#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
#define M 20
#define N 10000

struct P{
  char a[M];
  int key;
}e[N];

bool operator < (P a,P b){
  return a.key > b.key;
}

ll n;

int main(){
  scanf("%lld",&n);
  for(int i = 1;i <= n;++i)
  	scanf("%s%d",e[i].a + 1,&e[i].key);
  std::sort(e + 1,e + n + 1);
  std::cout<<e[2].a + 1<<std::endl;
}

(C)
考虑到只有四位数，所以直接枚举并判断是否合法就行。

C

#include<iostream>
#include<cstdio>
#define ll long long

char ai[20];

ll ans;

inline bool check(ll now){
	ll a = now / 1000,b = now % 1000 / 100,c = now % 100 / 10,d = now % 10;
	for(int i = 0;i <= 9;++i)
	if(ai[i] == 'o' && a != i && b != i && c != i && d != i)
	return false;
	else
	if(ai[i] == 'x' && (a == i || b == i || c == i || d == i))
	return false;
	return true;
}

int main(){
	scanf("%s",ai);
	for(int i = 0;i <= 9999;++i){
		if(check(i))
		ans ++ ;
	}
	std::cout<<ans<<std::endl;
}

(D)
没想到(ABC)还考(min-max)对抗搜索，考虑记录的答案是第一手比第二手高多少分就行了。（据说正解是反着(dp)）

D

#include<iostream>
#include<cstdio>
#define ll long long
#define N 2005

ll h,w;
ll p[2005][2005],f[N][N];
bool v[N][N];
ll x,y;

inline ll dfs(int turn){
	if(v[x][y])return f[x][y];
	ll ans = turn ? -0x7f7f7f7f : 0x7f7f7f7f;
	if(x + 1 <= h){
		x += 1;
		if(turn)
		ans = std::max(ans,dfs(0) + p[x][y]);
		else
		ans = std::min(ans,dfs(1) - p[x][y]);
		x -= 1;
	}
	if(y + 1 <= w){
		y += 1;
		if(turn)
		ans = std::max(ans,dfs(0) + p[x][y]);
		else
		ans = std::min(ans,dfs(1) - p[x][y]);
		y -= 1;
	} 
//	std::cout<<turn<<" "<<x<<" "<<y<<" "<<ans<<std::endl; 
	v[x][y] = 1; 
	return f[x][y] = ans;
}

int main(){
	scanf("%lld%lld",&h,&w);
	for(int i = 1;i <= h;++i)
	for(int j = 1;j <= w;++j){
		char a;
		while(a != '+' && a != '-')
		a = getchar();
		if(a == '+')
		p[i][j] = 1;
		else
		p[i][j] = -1;
		a = 'q';
	}
	x = 1,y = 1;
	v[h][w] = 1,f[h][w] = 0;
	ll k = dfs(1);
//	std::cout<<k<<std::endl;
	if(k > 0)
	puts("Takahashi");
	if(k == 0)
	puts("Draw");
	if(k < 0)
	puts("Aoki");
}

(E)
考虑这种树上数对统计答案问题，是可以换根做的。考虑在某个点((x))为根时，求出答案为(sum_i dis(x,i))就行了。
对于换根操作，考虑到要换到的点到根的异或距离为(k)，则他到每个点的距离都要异或(k),
考虑分位，单位算贡献就好了。

E

#include<iostream>
#include<cstdio>
#define ll long long
#define N 200005
#define mod 1000000007

ll head[N],cnt;

ll n,dis[N],c[N],a[N];

struct P{int to,next;ll wi;}e[N * 2];

inline void add(int x,int y,ll w){
	e[++cnt].to = y;
	e[cnt].next = head[x];
	e[cnt].wi = w;
	head[x] = cnt;
}

inline void dfs1(int u,int f){
	for(int i = head[u];i;i = e[i].next){
		int v = e[i].to;
		if(v == f)
		continue;
		dis[v] = dis[u] ^ e[i].wi;
		dfs1(v,u);
	}
}

ll ans = 0;

inline void dfs2(int u,int f){
	ll k;
	for(int i = 0;i < 60;++i){
		if(dis[u] & (1ll << i))k = n - c[i];
		else
		k = c[i];
		ans += (1ll << i) % mod * k % mod;
		ans %= mod;
	}
	for(int i = head[u];i;i = e[i].next){
		int v = e[i].to;
		if(v == f)
		continue;
		dfs2(v,u);
	}
}

int main(){
	scanf("%lld",&n);
	for(int i = 1;i <= n - 1;++i){
		ll u,v,w;
		scanf("%lld%lld%lld",&u,&v,&w);
		add(u,v,w);
		add(v,u,w);
	}
	dfs1(1,0);
	for(int i = 1;i <= n;++i)
	for(int j = 0;j < 60;++j)
	if(dis[i] & (1ll << j))c[j] ++ ;
	dfs2(1,0);
	std::cout<<ans * (500000004) % mod<<std::endl;
}