“美登杯”上海市高校大学生程序设计邀请赛补题

“美登杯”上海市高校大学生程序设计邀请赛补题

传送门

A题

题目吓人，实际上就是求子串的个数。长度为n的字符串子串个数为(n+1)*n/2

#include<bits/stdc++.h>
using namespace std;
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define scd(x) scanf("%lf",&x)
#define scd2(x,y) scanf("%lf%lf",&x,&y)
#define scd3(x,y,z) scanf("%lf%lf%lf",&x,&y,&z)
#define scc(x) scanf("%c",&x)
#define scs(x) scanf("%s",x)
#define pri(x) printf("%d
",x)
#define pri2(x,y) printf("%d %d
",x,y)
#define pri3(x,y,z) printf("%d %d %d
",x,y,z)
#define prl(x) printf("%lld
",x)
#define prl2(x,y) printf("%lld %lld
",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld
",x,y,z)
#define prc(x) printf("%c",x)
#define prd(x) printf("%lf
",x)
#define prt(x) printf("%s",(x))
#define prs(x) printf("%s
",(x))
#define ll long long
#define LL long long
#define ULL unsigned long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PDD pair<double,double>
#define pi acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define N 205
const int mod = 998244353;
const int P = 1e9 + 7;
const int maxn = 1e4+5;
int n,q;
ll l,r;
char s[maxn];
int main()
{
  sca2(n,q);
  scs(s);
  for(;q;q--)
  {
    scl2(l,r);
    prl((r-l+1)*(r-l+2)/2);
  }
}

B题

三重循环暴力求三角形。除了正三角形，还有倒三角形

#include<bits/stdc++.h>
using namespace std;
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define scd(x) scanf("%lf",&x)
#define scd2(x,y) scanf("%lf%lf",&x,&y)
#define scd3(x,y,z) scanf("%lf%lf%lf",&x,&y,&z)
#define scc(x) scanf("%c",&x)
#define scs(x) scanf("%s",x)
#define pri(x) printf("%d
",x)
#define pri2(x,y) printf("%d %d
",x,y)
#define pri3(x,y,z) printf("%d %d %d
",x,y,z)
#define prl(x) printf("%lld
",x)
#define prl2(x,y) printf("%lld %lld
",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld
",x,y,z)
#define prc(x) printf("%c",x)
#define prd(x) printf("%lf
",x)
#define prt(x) printf("%s",(x))
#define prs(x) printf("%s
",(x))
#define ll long long
#define LL long long
#define ULL unsigned long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PDD pair<double,double>
#define pi acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define N 205
const int mod = 998244353;
const int P = 1e9 + 7;
const int maxn = 1e5+5;
char a[N][N];
int vis[26][26][26];
int n;
int ans;
map<string,int>p;
string s;
int main()
{
  sca(n);
  rep(i,0,n+1)scs(a[i]);
  ans = 0;
  rep(i,0,n+1)
  {
    rep(j,0,i+1)
    {
      rep(k,1,n-i+1)
      {
        s.clear();
        s+=a[i][j];
        s+=a[i+k][j];
        s+=a[i+k][j+k];
        sort(s.begin(),s.end());
        if(!p[s])
        {
          p[s] = 1;
          ans++;
        }
      }
    }
  }
  for(int i=n;i>=0;i--)
  {
    for(int j=0;j<=i;j++)
    {
      for(int k=1;k<=i&&k<=j&&k<=i-j;k++)
      {
        s.clear();
        s+=a[i][j];
        s+=a[i-k][j];
        s+=a[i-k][j-k];
      //  cout << s<<endl;
        sort(s.begin(),s.end());
        if(!p[s])
        {
          p[s] = 1;
          ans++;
          //cout<<s<<endl;
        }
      }
    }
  }
  pri(ans);
}

C题

dfs进行连通块染色，两个点在k个图中都连通，那么这两个点所染的颜色序列是相同的
用map<vector,int> 存下每个点在k张图的颜色序列出现次数

#include<bits/stdc++.h>
using namespace std;
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define scd(x) scanf("%lf",&x)
#define scd2(x,y) scanf("%lf%lf",&x,&y)
#define scd3(x,y,z) scanf("%lf%lf%lf",&x,&y,&z)
#define scc(x) scanf("%c",&x)
#define scs(x) scanf("%s",x)
#define pri(x) printf("%d
",x)
#define pri2(x,y) printf("%d %d
",x,y)
#define pri3(x,y,z) printf("%d %d %d
",x,y,z)
#define prl(x) printf("%lld
",x)
#define prl2(x,y) printf("%lld %lld
",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld
",x,y,z)
#define prc(x) printf("%c",x)
#define prd(x) printf("%lf
",x)
#define prt(x) printf("%s",(x))
#define prs(x) printf("%s
",(x))
#define ll long long
#define LL long long
#define ULL unsigned long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PDD pair<double,double>
#define pi acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define N 205
const int mod = 998244353;
const int P = 1e9 + 7;
const int maxn = 1e6+5;
int n,m,k;
int u,v,color;
int vis[maxn];
vector<int>G[maxn],col[maxn];
map<vector<int>,int> p;
void dfs(int u)
{
  vis[u] = 1;
  col[u].push_back(color); 
  for(int i=0;i<G[u].size();i++) //对连通块进行染色
  {
    if(!vis[G[u][i]])
      dfs(G[u][i]);
  }
}
int main()
{
  sca2(n,k);
  while(k--)
  {
    memset(vis,0,sizeof vis);
    rep(i,0,n+1)
      G[i].clear();
    sca(m);
    while(m--)
    {
      sca2(u,v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    color = 0;
    rep(i,1,n+1)
    {
      if(!vis[i])
      {
        color++; // 新颜色
        dfs(i);
      }
    }
  }
  rep(i,1,n+1)
    p[col[i]]++;  //遍历每个点的颜色序列，拥有相同颜色序列的就是在k张图中都是连通的点
  rep(i,1,n+1)
    pri(p[col[i]]); 
}

D题

1.首先，如果每堆石子数量都为1，那么堆数为奇，先手胜，堆数为偶，先手负
2.如果第一堆石子是1，那么先手只能取这1个，相当于先手变成了“后手”。但如果下一堆还是1，那么又变回了先手。所以和1连续的堆数有关，连续堆数为偶，先手胜，连续堆数为奇，后手胜。
3.如果第一堆石子不是1，先手必胜。因为先手可以任意选择成为“先后手”，达到2的必胜态。

#include<bits/stdc++.h>
using namespace std;
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define scd(x) scanf("%lf",&x)
#define scd2(x,y) scanf("%lf%lf",&x,&y)
#define scd3(x,y,z) scanf("%lf%lf%lf",&x,&y,&z)
#define scc(x) scanf("%c",&x)
#define scs(x) scanf("%s",x)
#define pri(x) printf("%d
",x)
#define pri2(x,y) printf("%d %d
",x,y)
#define pri3(x,y,z) printf("%d %d %d
",x,y,z)
#define prl(x) printf("%lld
",x)
#define prl2(x,y) printf("%lld %lld
",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld
",x,y,z)
#define prc(x) printf("%c",x)
#define prd(x) printf("%lf
",x)
#define prt(x) printf("%s",(x))
#define prs(x) printf("%s
",(x))
#define ll long long
#define LL long long
#define ULL unsigned long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PDD pair<double,double>
#define pi acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define N 205
const int mod = 998244353;
const int P = 1e9 + 7;
const int maxn = 1e6+5;
int n,q;
int a[maxn],tot[maxn];
int cnt;
int main()
{
  sca(n);
  cnt=0;
  rep(i,1,n+1)
  {
    sca(a[i]);
    a[i+n] = a[i];
    if(a[i] == 1)
      cnt++;
  }
  for(int i=n+n;i>=1;i--)
  {
    if(a[i] == 1)
      tot[i] = tot[i+1]+1;
    else
      tot[i] = 0;
  }
  rep(i,1,n+1)
  {
    if(cnt==n)
       n%2?prs("First"):prs("Second");
    else
    {
      if(a[i]!=1)
        prs("First");
      else
        tot[i]%2==0?prs("First"):prs("Second");
    }
  }
}
 

I题

签到水题

#include<bits/stdc++.h>
using namespace std;
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define scd(x) scanf("%lf",&x)
#define scd2(x,y) scanf("%lf%lf",&x,&y)
#define scd3(x,y,z) scanf("%lf%lf%lf",&x,&y,&z)
#define scc(x) scanf("%c",&x)
#define scs(x) scanf("%s",x)
#define pri(x) printf("%d
",x)
#define pri2(x,y) printf("%d %d
",x,y)
#define pri3(x,y,z) printf("%d %d %d
",x,y,z)
#define prl(x) printf("%lld
",x)
#define prl2(x,y) printf("%lld %lld
",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld
",x,y,z)
#define prc(x) printf("%c",x)
#define prd(x) printf("%lf
",x)
#define prt(x) printf("%s",(x))
#define prs(x) printf("%s
",(x))
#define ll long long
#define LL long long
#define ULL unsigned long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PDD pair<double,double>
#define pi acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define N 205
const int mod = 998244353;
const int P = 1e9 + 7;
const int maxn = 1e3+5;
int a,b,c,d,n;
int v[N];
int main()
{
  sca(n),sca2(a,b),sca2(c,d);
  int sum = 0;
  rep(i,0,n)
  {
    sca(v[i]);
    sum+=v[i];
  }
  int ans = sum;
  if(sum >= a)
  {
    ans = sum - b;
  }
  if(sum >= c)
  {
    ans = min(ans,sum - d);
  }
  pri(ans);
}