20201126模拟

T1

solution

把字符串排序后插入到Trie树中，前面出现过几次相同形式的字符串，答案就加上几

code

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int b=131,P=1000429,maxn = 7e4+10,maxm = 1e7+10;
int n,vis[maxm];
char a[maxn];
long long ans;
int trie[maxn][30],cnt;
int build(char *c){
	int len = strlen(c+1),root = 0;
	for (int i = 1;i <= len;i++){
		int nxt = c[i]-'a';
		if (!trie[root][nxt]) trie[root][nxt] = ++cnt;
		root = trie[root][nxt];
	}	
	vis[root]++;
	return vis[root]-1;
}
int main(){
	freopen("anagram.in","r",stdin);
	freopen("anagram.out","w",stdout);
	scanf ("%d",&n);
	for (int i = 1;i <= n;i++){
		scanf ("%s",a+1);
		int len = strlen(a+1);sort(a+1,a+len+1);
		ans += build(a);
		printf("%d
",ans);
	}
}

T2

solution

预处理出两两点之间的距离，跟(floyd)差不多，枚举中转点累加答案，水过80分

code

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
int read(){
	int x = 1,a = 0;char ch = getchar();
	while (ch < '0'||ch > '9'){if (ch == '-') x = -1;ch = getchar();}
	while (ch >= '0'&&ch <= '9'){a = a*10+ch-'0';ch = getchar();}
	return x*a;
}
const int maxn = 1005;
int n;
struct node{
	int to,nxt;
}ed[maxn];
int head[maxn],tot;
void add(int u,int to){
	ed[++tot].to = to;
	ed[tot].nxt = head[u];
	head[u] = tot;
}
int dep[maxn],f[maxn][30],dis[maxn][maxn];
void dfs(int x,int fa){
	dep[x] = dep[fa]+1;
	for (int i = 1;i <= 22;i++) f[x][i] = f[f[x][i-1]][i-1];
	for (int i = head[x];i;i = ed[i].nxt){
		int to = ed[i].to;
		if (to == fa) continue;
		f[to][0] = x;
		dfs(to,x);
	}
}
int lca(int x,int y){
	if (dep[x] > dep[y]) swap(x,y);
	for (int i = 22;i >= 0;i--){
		if (dep[f[y][i]] >= dep[x]) y = f[y][i]; 
	}
	if (x == y) return x;
	for (int i = 22;i >= 0;i--){
		if (f[x][i] != f[y][i]) x = f[x][i],y = f[y][i];
	}
	return f[x][0];
}
int main(){
	freopen("dist.in","r",stdin);
	freopen("dist.out","w",stdout);
	n = read();
	for (int i = 1;i <= n-1;i++){
		int x = read(),y = read();
		add(x,y),add(y,x);
	}
	dfs(1,0);
	for (int i = 1;i <= n;i++){
		for (int j = 1;j <= n;j++){
			if (i == j) continue;
			int tmp = lca(i,j);
			dis[i][j] = dep[i]+dep[j]-2*dep[tmp];
		}
	}
	int ans = 0;
	for (int i = 1;i <= n;i++){
		for (int j = 1;j <= n;j++){
			for (int k = 1;k <= n;k++){
				if (i == j||i == k||k == j) continue;
				ans += min(dis[i][j],min(dis[i][k],dis[k][j]));
			}
		}
	}
	printf("%d
",ans/6);
	return 0;
}

T3

20分dp：

状态：(dp_{i,j})表示(i)个数，或和为(j)的方案数

转移：(dp_{i,jmid k})+=(dp_{i-1,j})

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#define int long long
using namespace std;
int read(){
	int x = 1,a = 0;char ch = getchar();
	while (ch < '0'||ch > '9'){if (ch == '-') x = -1;ch = getchar();}
	while (ch >= '0'&&ch <= '9'){a = a*10+ch-'0';ch = getchar();}
	return x*a;
}
const int maxn = 1e3+10,mod = 998244353;
int n,m,p;
int dp[maxn][maxn];
int qpow(int a,int k){
	int res = 1;a = a%mod;
	while (k){
		if (k&1) res = res*a%mod;
		a = a*a%mod;
		k >>= 1;
	}
	return res;
}
int getnum(int x){
	int res = 0;
	while (x) res++,x -= x&(-x);
	return res;
}
signed main(){
	freopen("or.in","r",stdin);
	freopen("or.our","w",stdout);
	n = read(),m = read(),p = read();
	if (n <= 100&&m <= 100){
		for (int i = 0;i <= m;i++) dp[1][i] = 1;
		for (int i = 2;i <= n;i++){
			for (int j = 0;j < 2*m;j++){
				for (int k = 0;k <= m;k++){
					(dp[i][j|k] += dp[i-1][j])%=mod;
				}
			}
		}
		int ans = 0;
		for (int i = 0;i < 2*m;i++){
			if (dp[n][i]) (ans += qpow(p,i)*dp[n][i]%mod)%=mod;
		}
		printf("%lld
",ans);
		return 0;
	}
	int ans = 0;
	for (int i = 1;i <= m;i++){
		int res = getnum(i);
		int num = qpow((qpow(2,n)-1),res)%mod;
		(ans += qpow(p,i)*num%mod) %= mod;
	}
	printf("%lld
",ans);
	return 0;
}

T4

solution

dfs枚举全排列，看是否合法，合法再统计答案 10分

code

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
int read(){
	int x = 1,a = 0;char ch = getchar();
	while (ch < '0'||ch > '9'){if (ch == '-') x = -1;ch = getchar();}
	while (ch >= '0'&&ch <= '9'){a = a*10+ch-'0';ch = getchar();}
	return x*a;
}
const int maxn = 100;
int n,k;
bool vis[maxn];
int a[maxn],sum[maxn],num[maxn];
void fix(int x,int k){
    for (int i = x;i <= n;i+=i&(-i)) sum[i] += k;
}
int query(int x){
    int res = 0;
    for (int i = x;i >= 1;i-=i&(-i)) res += sum[i];
    return res;
}
void dfs(int x){
    if (x == n+1){
        for (int i = 1;i <= n;i++){
            if (a[a[i]] != i) return;
        }
        int ans = 0;
        memset(sum,0,sizeof(sum));
        for(int i = n;i >= 1;i--){
        	fix(a[i],1);
        	ans += query(a[i]-1);
        }
   		num[ans]++;
    }
    for (int i = 1;i <= n;i++){
        if (vis[i] == 0){
        	vis[i] = 1,a[x] = i;
        	dfs(x+1);
        	vis[i] = 0;
    	}
	}
}
int main(){
	freopen("perm.in","r",stdin);
	freopen("perm.out","w",stdout);
	n = read(),k = read();
	dfs(1);
	for (int i = 0;i <= k-1;i++) printf("%d",num[i]%2);
	return 0;
}