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A层省选6

作者：互联网

A. T1

考虑每一位对\(f\)的贡献，假设有\(x\)个\(a_i\)该位为\(1\)

code 
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
	ll x = 0; char c = getchar();
	while(c < '0' || c > '9')c = getchar();
	do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c <= '9' && c >= '0');
	return x;
}
const int maxn = 100005;
ll n, l, r;
ll work(){
	if(l == r)return 1;
	int hi = 0;  for(int i = 63; i >= 0; --i)if(((1ll << i) & l) != ((1ll << i) & r)){hi = i; break;}
	int low = -1; for(int i = hi - 1; i >= 0; --i)if(r & (1ll << i)){low = i; break;}
	ll nans = (1ll << hi) - (l & ((1ll << hi) - 1)), rl = (1ll << hi) - nans;
	ll lr = (1ll << (low + 1)) - 1;
	nans = nans << 1;
	return nans + min(lr, rl - 1) + 1;
}
int main(){
	freopen("a.in", "r", stdin);
	freopen("a.out", "w", stdout);
	int T = read();
	for(register int ask = 1; ask <= T; ++ask){
		n = read(), l = read(), r = read();
		if(n > 1)printf("%lld\n",work());
		else printf("%lld\n", r - l + 1);
	}
	return 0;
}

B. T2

考场刚推出对称差虎哥就发解释了..

这题先说对于任意四元组，在一棵树上有且仅有一种划分

划分在同一侧的两组在树上路径没有交，中间有一条链相连

于是我们可以枚举删去一条边或者一个点来容斥

由于一个四元组在边上统计的次数比在点上统计的次数多\(1\)，所以加上删边的贡献，减去删点的贡献

然后我们枚举在第一棵树上删了啥，对删除后在不同块的树点进行染色，在第二棵树上统计交集（枚举删去点/边，选所有不同颜色的组合，分别在不同子树内选取某种颜色的二元组），根据点/边决定是加上还是减去

这样$ S \Delta S^,= 2 \times C_{leaf}^4 - ans$

uglycode 
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<assert.h>
using namespace std;
typedef long long ll;
inline int read(){
	int x = 0; char c = getchar();
	while(c < '0' || c > '9')c = getchar();
	do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c <= '9' && c >= '0');
	return x;
}
const int maxn = 50005;
vector<int>g1[maxn], g2[maxn];
int leaf, n, col;
int cc[5], f[maxn][5];

ll ans, nans;
//edge - node = 1
int select2(int x){
	// assert(x >= 0);
	return x * (x - 1) / 2;
}
void solve(int x, int fa){
	// if(x <= leaf)return;
	int ls = 0, rs = 0;
	for(int v : g2[x]){
		if(v == fa)continue;
		solve(v, x);
		f[x][1] += f[v][1];
		f[x][2] += f[v][2];
		f[x][3] += f[v][3];
		if(ls)rs = v; else ls = v;
	}
	if(!ls || !rs)return;
	// assert((ls == 0 && rs == 0) || (ls != 0 && rs != 0));
	if(fa){
		int f1 = cc[1] - f[x][1], f2 = cc[2] - f[x][2], f3 = cc[3] - f[x][3];		
		nans = nans - select2(f1) * (0ll + select2(f[ls][2]) + select2(f[ls][3]) + select2(f[rs][2]) + select2(f[rs][3]));
		nans = nans - select2(f2) * (0ll + select2(f[ls][1]) + select2(f[ls][3]) + select2(f[rs][1]) + select2(f[rs][3]));
		nans = nans - select2(f3) * (0ll + select2(f[ls][1]) + select2(f[ls][2]) + select2(f[rs][1]) + select2(f[rs][2]));
		// printf("tree 2 node : %d del node now nans = %lld\n", x, nans);
	}
	nans = nans - select2(f[ls][1]) * (0ll + select2(f[rs][2]) + select2(f[rs][3]));
	nans = nans - select2(f[ls][2]) * (0ll + select2(f[rs][1]) + select2(f[rs][3]));
	nans = nans - select2(f[ls][3]) * (0ll + select2(f[rs][1]) + select2(f[rs][2]));
	// printf("tree 2 node : %d del node now nans = %lld\n", x, nans);

	nans = nans + select2(f[ls][1]) * (0ll + select2(cc[2] - f[ls][2]) + select2(cc[3] - f[ls][3]));
	nans = nans + select2(f[ls][2]) * (0ll + select2(cc[1] - f[ls][1]) + select2(cc[3] - f[ls][3]));
	nans = nans + select2(f[ls][3]) * (0ll + select2(cc[1] - f[ls][1]) + select2(cc[2] - f[ls][2]));
	// printf("tree 2 node : %d del edge - > %d now nans = %lld\n", x, ls,  nans);
	
	nans = nans + select2(f[rs][1]) * (0ll + select2(cc[2] - f[rs][2]) + select2(cc[3] - f[rs][3]));
	nans = nans + select2(f[rs][2]) * (0ll + select2(cc[1] - f[rs][1]) + select2(cc[3] - f[rs][3]));
	nans = nans + select2(f[rs][3]) * (0ll + select2(cc[1] - f[rs][1]) + select2(cc[2] - f[rs][2]));
	// printf("%d %d %d %d %d %d\n",f[rs][1], f[rs][2], f[rs][3], cc[1], cc[2], cc[3]);
	// printf("tree 2 node : %d del edge - > %d now nans = %lld\n", x, rs, nans);

}

void upd(int x, int fa){
	for(int i = 1; i <= 4; ++i)f[x][i] = 0;
	// printf("upd : %d %d %d\n", x, fa, col);
	if(x <= leaf){ ++f[x][col]; ++cc[col]; return; }
	for(int v : g1[x]) if(v != fa)upd(v, x);
}
void del(int x, int fa){
	col = 0; cc[1] = cc[2] = cc[3] = 0;
	for(int v : g1[x]){
		for(int i = 1; i <= 4; ++i)f[x][i] = 0;
		++col, upd(v, x);
	}	
	nans = 0; solve(1, 0); ans = ans - nans; 
	// printf("del node : %d now nans = %lld\n\n",x , nans);
	for(int v : g1[x]){
		if(v == fa)continue;
		cc[1] = cc[2] = cc[3] = 0;
		col = 1; upd(v, x); col = 2; upd(x, v);
		nans = 0; solve(1, 0); ans = ans + nans;
		// printf("del edge : %d < - > %d now nans = %lld\n\n",x, v, nans);
	}
	for(int v : g1[x])if(v != fa)del(v, x);
}
int main(){
	freopen("b.in","r",stdin);
	freopen("b.out","w",stdout);
	leaf = read(); n = leaf + leaf - 2;
	for(int i = 1; i < n; ++i){
		int u = read(), v = read();
		g1[u].push_back(v), g1[v].push_back(u);
	}
	for(int i = 1; i < n; ++i){
		int u = read(), v = read();
		g2[u].push_back(v), g2[v].push_back(u);	
	}
	del(n, 0);
	ans = 1ll * leaf * (leaf - 1) * (leaf - 2) * (leaf - 3) / 12 - ans - ans;
	printf("%lld\n",ans);
	return 0;
}

C. T3

标签：层省,rs,int,cc,nans,include,select2
来源： https://www.cnblogs.com/Chencgy/p/16597459.html