2018 计蒜之道 复赛

A. 贝壳找房函数最值

常规贪心推式子。按(a-1)/b排序

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define frep(i,a,b) for(int i=a;i>=b;--i)
#define mem(W) memset(W,0,sizeof(W))
#define pb push_back
typedef long long ll;
const int N = 1e5+100;
const int inf = 0x3f3f3f3f;
inline int read() {char c=getchar(); int x=0,f=1;while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}return x*f;
}
using namespace std;
int n,x;
struct node{int a,b;} A[N];
bool cmp(node x,node y) {return (x.a-1)*y.b > (y.a-1)*x.b;
}
int main() {int T = read();while(T--){n = read(); x= read();rep(i,1,n)A[i].a=read();rep(i,1,n)A[i].b=read();sort(A+1,A+1+n,cmp);frep(i,n,1) x = (A[i].a*x%10+ A[i].b)%10;printf("%d\n",x);}return 0;
}

D. 贝壳找房魔法师顾问

情况1：两边都不可变。对比是否相等，直接判断；

情况2：两边都变。对于每个联通块找一颗生成树最优；

情况3：一边可变，另一边固定。对每个弱联通分量，如果是DAG，显然可以按拓扑序连成链即可保证连接关系，如果有环就把所有节点串成环，它们就可以完成任意的变化了。

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define frep(i,a,b) for(int i=a;i>=b;--i)
#define mem(W) memset(W,0,sizeof(W))
#define pb push_back
typedef long long ll;
const int N = 2e5+100;
const int inf = 0x3f3f3f3f;
inline int read() {char c=getchar(); int x=0,f=1;while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}return x*f;
}
using namespace std;
int n, a[N], b[N];
map<int,int> M[N],M2[N];
char s1[111],s2[111];
struct edge{int u,v,nxt;}E[N*2][2];
int cc,h[N][2];
void add(int u,int v) {++cc;E[cc][0].u=u;E[cc][0].v=v;E[cc][0].nxt=h[u][0];h[u][0]=cc;E[cc][1].u=v;E[cc][1].v=u;E[cc][1].nxt=h[v][1];h[v][1]=cc;
}
int vis[N];vector<int> s;
int dfs1(int u){vis[u] = 1;s.pb(u);int ans = 1;for(int i=h[u][0];~i;i=E[i][0].nxt) {int v = E[i][0].v;if(!vis[v])ans += dfs1(v);}for(int i=h[u][1];~i;i=E[i][1].nxt) {int v = E[i][1].v;if(!vis[v]) ans += dfs1(v);}return ans;
}
int vis1[N];
int in[N],q[N],tp;
int inc() {int n=s.size();tp = 0; rep(i,0,n-1) if(!in[s[i]]) q[tp++]=s[i];rep(i,0,tp-1) {for(int j=h[q[i]][0];~j;j=E[j][0].nxt) {--in[E[j][0].v];if(!in[E[j][0].v]) q[tp++]=E[j][0].v;}}
//    if(n==2){
//        rep(i,0,n-1)cout << s[i] << endl;
//    }s.clear();
//    if(n==2){
//        cout << tp << endl;
//    }return (n != tp);
}
int solve() {int ans = 0;rep(i,1,100000)if(!vis[i]){ans += (dfs1(i) - !(inc()));}return ans;
}int bfs(int s) {int ans = 0;vis[s]=1;++ans;queue<int> q;q.push(s);while(!q.empty()){int u=q.front();q.pop();for(int i=h[u][0];~i;i=E[i][0].nxt){int v = E[i][0].v;if(!vis[v]){vis[v]=1;++ans;q.push(v);}}}return ans;
}int main() {n = read();//int m=read();memset(h,-1,sizeof(h));
//    rep(i,1,m){int u,v;
//        scanf("%d%d",&u,&v);
//        add(u,v);
//        ++in[v];
//    }
//    rep(i,1,n)s.pb(i);
//    inc();scanf(" %s",s1);rep(i,1,n)scanf("%d",&a[i]);scanf(" %s",s2);rep(i,1,n)scanf("%d",&b[i]);if(s1[0]=='C'&&s2[0]=='C'){int f=0;rep(i,1,n)if(a[i]!=b[i]){f=1;break;}if(f)puts("-1");else puts("0");}else {if(s1[0]=='C'&&s2[0]!='C'){int ans=0;rep(i,1,n)if(a[i]!=b[i]&&!M[b[i]][a[i]]){add(b[i],a[i]); ++in[a[i]];M[b[i]][a[i]]=1;}printf("%d\n",solve());}else if(s1[0]!='C'&&s2[0]=='C'){int ans=0;rep(i,1,n)if(a[i]!=b[i]&&!M[a[i]][b[i]]){add(a[i],b[i]); ++in[b[i]];M[a[i]][b[i]] = 1;}printf("%d\n",solve());}else {rep(i,1,n)if(a[i]!=b[i]&&!M[a[i]][b[i]]){add(a[i],b[i]);add(b[i],a[i]);M[a[i]][b[i]] = M[b[i]][a[i]] = 1;}int ans=0;rep(i,1,100000)if(!vis[i])ans+=bfs(i)-1;printf("%d\n",ans);}}return 0;
}

G. 贝壳找房计数比赛

字符串的排列数 = (长度)!/(a的数目!)/.../(z的数目!)，先把t挑出来。剩下的放到两边，就相当于把剩余字符的全排列，从一个位置分成两块，有(n+1)个位置。

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define frep(i,a,b) for(int i=a;i>=b;--i)
#define mem(W) memset(W,0,sizeof(W))
typedef long long ll;
const int N = 1e5+100;
const int inf = 0x3f3f3f3f;
const ll mod = 1e9 + 7;
inline int read() {char c=getchar(); int x=0,f=1;while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}return x*f;
}
using namespace std;
int n,a[N],A[27];
string s1,s2;
int p[N], notp[N], nxt[N], b[N];
void init(){notp[1]=1;nxt[1]=1;for(int i=2;i<=100000;++i) {if(!notp[i]) p[++p[0]]=i,nxt[i]=i;for(int j=1;j<=p[0]&&i*p[j]<=100000;++j) {notp[i*p[j]] = 1;nxt[i*p[j]]=p[j];if(i%p[j]==0)break;}}
}
ll q_pow(ll a,ll b) {ll ans=1;while(b) {if(b&1LL) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1LL;}return ans;
}
inline void add(int x,int f) {while(x!=1){b[nxt[x]]+=f;x/=nxt[x];}
}
ll cal() {ll ans = 1;rep(i,1,p[0])ans = (ans*q_pow(p[i],b[p[i]]))%mod;return ans%mod;
}
int main() {init();int T = read();while(T--) {int f=0, num=0;cin >> s1 >> s2;mem(A);mem(b);int n1=s1.size(), n2=s2.size();rep(i,0,n1-1) ++A[s1[i]-'a'];rep(i,0,n2-1) --A[s2[i]-'a'];rep(i,0,25){if(A[i]<0){puts("0");f=1;break;}num += A[i];}if(f)continue;rep(i,1,num) add(i,1);rep(i,0,25) rep(j,1,A[i]) add(j,-1);printf("%lld\n", (cal()*(num+1))%mod);}return 0;
}