2021.02.18 北师大寒假新生训练

Candies and Two Sisters

签到

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main() {int T;scanf("%d", &T);while (T--) {int n;scanf("%d", &n);if (n & 1) printf("%d\n", n / 2);else printf("%d\n", n / 2 - 1);}return 0;
}

Construct the String

从’a’ 到 ‘a’ + b 循环往里面填字母就好。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 2010;char s[maxn];int main() {int T;scanf("%d", &T);while (T--) {int n, a, b;scanf("%d%d%d", &n, &a, &b);for (int i = 0; i < n; i++) {s[i] = 'a' + i % b;}s[n] = 0;printf("%s\n", s);}return 0;
}

Two Teams Composing

分类讨论题
{type,type≤m−1;m−1,type=m;m,type>m.\begin{cases}type, \quad type \le m-1; \\ m-1, \quad type = m; \\ m, \quad type > m.\end{cases}⎩⎪⎨⎪⎧type,type≤m−1;m−1,type=m;m,type>m.

#include<iostream>
#include<algorithm>
#include<cstring>
#include<unordered_map>
using namespace std;unordered_map<int, int> um;
int main() {int T;scanf("%d", &T);while (T--) {int n;scanf("%d", &n);um.clear();int m = 0;for (int i = 0; i < n; i++) {int x;scanf("%d", &x);um[x]++;m = max(um[x], m);}int type = um.size();if (type <= m - 1) printf("%d\n", type);else if (type == m) printf("%d\n", m - 1);else printf("%d\n", m);}return 0;
}

Anti-Sudoku

依次挑选小矩阵中的元素，比如挑选第(i,j) 的小矩形，那么就挑选小矩形中的 (j,i) 元素，只要修改的和之前的不一样，那么一定就会满足条件，因为该行该列该小矩阵都是只有这一个元素被修改掉。

#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;char mz[15][15];int main() {int T;scanf("%d", &T);while (T--) {for (int i = 1; i <= 9; i++) scanf("%s", mz[i] + 1);for (int i = 1; i <= 3; i++) {for (int j = 1; j <= 3; j++) {int x = 3 * (j - 1) + i, y = 3 * (i - 1) + j;if (mz[x][y] == '1') mz[x][y] = '2';else mz[x][y] = '1';}}for (int i = 1; i <= 9; i++) printf("%s\n", mz[i] + 1);}return 0;
}

Three Blocks Palindrome (hard version)

我们发现a的范围很小，于是我们可以先确定第一个数字，然后用双指针确定x的值，那么中间y的值就是两边对称的x，所夹区间出现次数最多的数字是多少。
那么难点就在于，我们能在 O(1)O(1)O(1) 求数组中某个区间，某个数字出现的次数。
其实可以用前缀和，开一个数组 a[200010][210]a[200010][210]a[200010][210], a[i][j]a[i][j]a[i][j] 记录数字 jjj 在第 i 个位置是否出现。出现的话为1，没有出现为0。然后求这个数字的前缀和。这样子，就解决了上述那个问题。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;const int maxn = 200010, maxa = 210;
int a[maxn][maxa];
vector<int> ids[maxa];int main() {int T;scanf("%d", &T);while (T--) {int N;scanf("%d", &N);for (int i = 1; i <= 200; i++) ids[i].clear();for (int i = 1; i <= N; i++) {memset(a[i], 0, sizeof a[i]);int x;scanf("%d", &x);a[i][x] = 1;ids[x].push_back(i);}int maxsz = 0;for (int j = 1; j <= 200; j++) {maxsz = max(maxsz, (int)ids[j].size());for (int i = 1; i <= N; i++) {a[i][j] += a[i - 1][j];}}int ans = maxsz;for (int m = 1; m <= 200; m++) {if ((int)ids[m].size() <= 1) continue;int sz = (int)ids[m].size();int l = 0, r = sz - 1;while (r > l) {int x = l + 1;int y = 0;for (int i = 1; i <= 200; i++) {y = max(y, a[ids[m][r] - 1][i] - a[ids[m][l]][i]);//if (i == 1) printf("### %d\n", a[ids[m][r] - 1][i] - a[ids[m][l]][i]);}ans = max(2 * x + y, ans);l++, r--;}}printf("%d\n", ans);}
}

Robots on a Grid

图论（队友写的）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<cmath>
#include<iostream>
using namespace std;
int n,m;
string dir[1000010],color[1000010];
int occupy[1000010];
int p[1000010][30],ans1,ans2;
void init()
{ans1=ans2=0;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){int x=(i-1)*m+j;occupy[x]=-1;for(int k=0;k<=25;k++)p[x][k]=0;}}for(int i=1;i<=n;i++){for(int j=0;j<m;j++){int u=(i-1)*m+j+1;if(dir[i][j]=='U'){p[u][0]=u-m;}else if(dir[i][j]=='D'){p[u][0]=u+m;}else if(dir[i][j]=='L'){p[u][0]=u-1;}else if(dir[i][j]=='R'){p[u][0]=u+1;}}}for(int j=1;(1<<j)<=n*m;j++){for(int i=1;i<=n*m;i++){p[i][j]=p[p[i][j-1]][j-1];}}
}
void solve()
{for(int i=1;i<=n;i++){for(int j=0;j<m;j++){int c=color[i][j]-'0',u=(i-1)*m+j+1;for(int k=(int)log2(n*m);k>=0;k--){if(!p[u][k])continue;u=p[u][k];}if(occupy[u]!=0)occupy[u]=c;}}for(int i=1;i<=n*m;i++){if(occupy[i]!=-1)ans1++;if(occupy[i]==0)ans2++;}
}
int main()
{int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){cin>>color[i];}for(int i=1;i<=n;i++){cin>>dir[i];}init();solve();printf("%d %d\n",ans1,ans2);}return 0;
}

Decimal

找找规律会发现，把n质因数分解后，只要分解的结果，含有除2或5以外其他质因数，那么就是无限小数。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int main() {int T;cin >> T;while (T--) {int n;cin >> n;while (n % 2 == 0) n /= 2;while (n % 5 == 0) n /= 5;if (n == 1) printf("No\n");else printf("Yes\n");}return 0;
}

Forest Program

图论

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod=998244353;
ll vis[300010],pre[300010],sum;
ll h[1000010],ne[1000010],e[1000010],idx;
ll ans=1;
ll qmi(ll m,ll k)
{ll res=1,t=m;while(k){if(k&1)res=res*t%mod;t=t*t%mod;k>>=1;}return res;
}
void add(ll a,ll b)
{e[++idx]=b;ne[idx]=h[a];h[a]=idx;
}
void dfs(ll u,ll fa)
{vis[u]=1;for(ll i=h[u];i;i=ne[i]){ll v=e[i];if(v==fa)continue;if(vis[v]==2)continue;if(!vis[v]){pre[v]=u;dfs(v,u);}else{ll step=1;ll tmp=u;while(tmp!=v){step++;tmp=pre[tmp];}sum+=step;ans=ans*(qmi(2,step)-1)%mod;}}vis[u]=2;
}
int main()
{ll n,m;scanf("%lld%lld",&n,&m);for(ll i=1;i<=m;i++){ll a,b;scanf("%lld%lld",&a,&b);add(a,b);add(b,a);}for(ll i=1;i<=n;i++){if(!vis[i]){dfs(i,0);}}ll res=m-sum;ans=ans*qmi(2,res)%mod;printf("%lld",ans);return 0;
}

Angle Beats

给定一个点，问有多少条点对，可以与这个点构成一个直角三角形。
思路很简单，就是非常卡时间。我们分为 AiA_iAi 是否作为直角边分类讨论。把所有的边存在哈希表中，然后查找。
省时间策略:

不用map，用unordered_map
unordered_map里面不要存pair，更不要存三元组，存一维的元素，省空间省时间。具体hash的办法很简单，就是 x * hash_num + y 即可。
把线段方向保存为向量(dx, dy)，而不是保存倾斜角，防止浮点数误差。dx, dy 都除以最大公约数，用这种方式把它们都标准化。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<unordered_map>
using namespace std;#define x first
#define y second
typedef long long ll;
const int maxn = 2010;
const ll hash_num = 3e9 + 9;
typedef pair<int, int> P;ll Hash(const P& p) {return hash_num * (ll)p.x + (ll)p.y;
}P ps[maxn], qs[maxn];
unordered_map<ll, int> tmap;
int ans[maxn];inline int gcd(int a, int b) {if (b == 0) return a;return gcd(b, a % b);
}inline void normalize(P& p) {int dx = p.x, dy = p.y;int d = gcd(abs(dx), abs(dy));dy /= d, dx /= d;p.x = dx, p.y = dy;
}int main() {int N, Q;scanf("%d%d", &N, &Q);for (int i = 1; i <= N; i++) {scanf("%d%d", &ps[i].x, &ps[i].y);}for (int i = 1; i <= Q; i++) {scanf("%d%d", &qs[i].x, &qs[i].y);}for (int i = 1; i <= Q; i++) {tmap.clear();for (int j = 1; j <= N; j++) {P p(qs[i].x - ps[j].x, qs[i].y - ps[j].y);normalize(p);tmap[Hash(p)]++;}for (int j = 1; j <= N; j++) {P p(qs[i].x - ps[j].x, qs[i].y - ps[j].y);normalize(p);//两个方向都要加上去ans[i] += tmap[Hash(P(-p.y, p.x))];ans[i] += tmap[Hash(P(p.y, -p.x))];}ans[i] /= 2;}for (int i = 1; i <= N; i++) {tmap.clear();for (int j = 1; j <= N; j++) {if (i != j) {P p = { ps[i].x - ps[j].x, ps[i].y - ps[j].y };normalize(p);tmap[Hash(p)]++;}}for (int j = 1; j <= Q; j++) {P p = { ps[i].x - qs[j].x, ps[i].y - qs[j].y };normalize(p);ans[j] += tmap[Hash(P(-p.y, p.x))];ans[j] += tmap[Hash(P(p.y, -p.x))];}}for (int i = 1; i <= Q; i++) {printf("%d\n", ans[i]);}return 0;
}

Invoker

dp问题，感觉应该归为一个状态机的问题。dp[i][j]dp[i][j]dp[i][j] = min(dp[i][j],dp[i−1][k]+dist(k→j)min(dp[i][j],dp[i-1][k]+dist(k \rightarrow j)min(dp[i][j],dp[i−1][k]+dist(k→j)。搜索从 i - 1 到 i 状态，从6种字母到另外 6 种字母，的最小值。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<unordered_map>
using namespace std;char words[10][6][4] = { {"QQQ","QQQ","QQQ","QQQ","QQQ","QQQ"},{"QQW","QWQ","WQQ","WQQ","WQQ","WQQ"},{"QQE","QEQ","EQQ","EQQ","EQQ","EQQ"},{"WWW","WWW","WWW","WWW","WWW","WWW"},{"QWW","WQW","WWQ","WWQ","WWQ","WWQ"},{"WWE","WEW","EWW","EWW","EWW","EWW"},{"EEE","EEE","EEE","EEE","EEE","EEE"},{"QEE","EQE","EEQ","EEQ","EEQ","EEQ"},{"WEE","EWE","EEW","EEW","EEW","EEW"},{"QWE","QEW","EQW","EWQ","WEQ","WQE"},
};const int maxn = 100010;
char str[maxn], p[maxn];
int f[maxn][6];
unordered_map<char, int> um;int dist(int a, int b, int x, int y) {//计算从 words[a][b] 到 words[x][y] 需要在末尾添加几个字母，即用几次技能。if (words[a][b][0] == words[x][y][0] && words[a][b][1] == words[x][y][1] && words[a][b][2] == words[x][y][2]) {return 0;}if (words[a][b][1] == words[x][y][0] && words[a][b][2] == words[x][y][1]) {return 1;}if (words[a][b][2] == words[x][y][0]) {return 2;}return 3;
}int main() {um['X'] = 0; um['V'] = 1; um['G'] = 2;um['C'] = 3; um['X'] = 4; um['Z'] = 5;um['T'] = 6; um['F'] = 7; um['D'] = 8; um['B'] = 9;scanf("%s", str);//前后两个字母相同只需要按R，把这些东西先挑出来。int sz = 0, res1 = strlen(str);p[sz++] = um[str[0]];for (int i = 1; str[i]; i++) {if (str[i] != str[i - 1]) p[sz++] = um[str[i]];}//for (int i = 0; i < sz; i++) {//    printf("* %d\n", p[i]);//}//printf("%d\n", res1);memset(f, 0x3f, sizeof f);for (int j = 0; j < 6; j++) f[0][j] = 3;for (int i = 1; i < sz; i++) {for (int j = 0; j < 6; j++) {for (int k = 0; k < 6; k++) {f[i][j] = min(f[i][j], f[i - 1][k] + dist(p[i - 1], k, p[i], j));}//printf("### %d %d %d\n", i, j, f[i][j]);}}int res2 = 1e9;for (int j = 0; j < 6; j++) {res2 = min(res2, f[sz - 1][j]);}printf("%d\n", res1 + res2);return 0;
}

MUV LUV EXTRA

字符串，KMP + 最小循环节。我们发现一个规律（假设ne数组和模式串下标均从1开始）：最小循环节（题目中的 l）= i - ne[i]. （其实告诉这个规律后，发现道理也挺显然的）
i 从 1 到 N 一直更新答案就可以了 ans = max(ans, a * i - b * (i - ne[i]))

完整代码：

#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;const int maxn = 10000010;
int ne[maxn], N;
char str[maxn], p[maxn];
typedef long long ll;
ll a, b;
int main() {scanf("%lld%lld%s", &a, &b, str);for (int i = 0; str[i]; i++) {if (str[i] == '.') {strcpy(p + 1, str + i + 1);break;}}N = strlen(p + 1);reverse(p + 1, p + N + 1);//计算 ne 数组for (int i = 2, j = 0; i <= N; i++) {while (j && p[i] != p[j + 1]) j = ne[j];if (p[i] == p[j + 1]) j++;ne[i] = j;}//printf("### %s\n", p + 1);//for (int i = 1; i <= N; i++) printf("*** %d %d\n", i, ne[i]);ll ans = -1e9;for (int i = 1; i <= N; i++) {ll p = i, l = i - ne[i];ans = max(ans, a * p - b * l);}printf("%lld\n", ans);return 0;
}

MUV LUV UNLIMITED

树上博弈

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int fa[1000010],du[1000010],n;
int h[1000010],ne[1000010],e[1000010],idx;
int depth[1000010];
void add(int a,int b)
{e[++idx]=b;ne[idx]=h[a];h[a]=idx;
}
void dfs(int u,int len)
{if(du[u]>1)len=0;depth[u]=len;for(int i=h[u];i;i=ne[i]){int v=e[i];dfs(v,len+1);}
}
int main()
{int T;scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=1;i<=n;i++){du[i]=0;h[i]=0;depth[i]=0;}idx=0;for(int i=2;i<=n;i++){int x;scanf("%d",&x);fa[i]=x;du[x]++;add(x,i);}bool flag=false;for(int i=1;i<=n&&!flag;i++){if(!du[i]){int x=fa[i];if(du[x]>1){flag=true;}}}for(int i=1;i<=n&&!flag;i++){if(!du[i]){if(depth[i]%2)flag=true;}}if(flag)printf("Takeru\n");elseprintf("Meiya\n");}return 0;
}

Escape

分层图 & 网络流

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int n,m,rob,exc;
int h[1000010],e[1000010],ne[1000010],capacity[1000010],flow[1000010],idx;
int S,T;
char cell[110][110];
int dis[60010],curedge[1000010];
void add(int a,int b,int c)
{e[idx]=b;ne[idx]=h[a];capacity[idx]=c;flow[idx]=0;h[a]=idx++;
}
void init()
{memset(h,-1,sizeof h);idx=0;
}
void input()
{scanf("%d%d%d%d",&n,&m,&rob,&exc);T=n*m*6;for(int i=1;i<=n;i++){scanf("%s",cell[i]+1);}for(int i=1;i<=rob;i++){int p;scanf("%d",&p);add(S,p,1);add(p,S,0);add(p,p+m,0x3f3f3f3f);add(p+m,p,0);}for(int i=1;i<=exc;i++){int p;scanf("%d",&p);int id=n*m*2+p;add(id,n*m*5+p,0x3f3f3f3f);add(n*m*5+p,id,0);add(n*m*5+p,T,0x3f3f3f3f);add(T,n*m*5+p,0);}
}
void build()
{for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(cell[i][j]=='1')continue;int up=i*m+j,down=up+n*m,left=up+n*m*2,right=up+n*m*3;add(up,down,1);add(down,up,0);add(left,right,1);add(right,left,0);add(down,left,0x3f3f3f3f);add(left,down,0);add(right,up,0x3f3f3f3f);add(up,right,0);if(cell[i-1][j]!='1'&&i-1>=1){add(down,(i-1)*m+j,0x3f3f3f3f);add((i-1)*m+j,down,0);}if(cell[i+1][j]!='1'&&i+1<=n){add(down,(i+1)*m+j,0x3f3f3f3f);add((i+1)*m+j,down,0);}if(cell[i][j-1]!='1'&&j-1>=1){add(right,left-1,0x3f3f3f3f);add(left-1,right,0);}if(cell[i][j+1]!='1'&&j+1<=m){add(right,left+1,0x3f3f3f3f);add(left+1,right,0);}}}
}
bool bfs()
{memset(dis,0x3f,sizeof dis);queue<int> q;q.push(S);dis[S]=0;while(q.size()){int u=q.front();q.pop();for(int i=h[u];i!=-1;i=ne[i]){int v=e[i];if(dis[v]==0x3f3f3f3f&&capacity[i]-flow[i]>0){dis[v]=dis[u]+1;q.push(v);}}}return dis[T]!=0x3f3f3f3f;
}
int dfs(int u,int limit)
{if(u==T)return limit;int delta=limit;for(int& i=curedge[u];i!=-1;i=ne[i]){int v=e[i];if(dis[v]==dis[u]+1&&capacity[i]-flow[i]>0){int d=dfs(v,min(delta,capacity[i]-flow[i]));flow[i]+=d;flow[i^1]-=d;delta-=d;}if(delta==0)break;}return limit-delta;
}
int dinic()
{int ans=0;while(bfs()){for(int i=0;i<=6*n*m;i++){curedge[i]=h[i];}ans+=dfs(S,0x3f3f3f3f);}return ans;
}
int main()
{int t;scanf("%d",&t);while(t--){init();input();build();int ans=dinic();if(ans==rob)printf("Yes\n");elseprintf("No\n");}return 0;
}

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2021.02.18 北师大寒假新生训练

2021.02.18 北师大寒假新生训练

Candies and Two Sisters

Construct the String

Two Teams Composing

Anti-Sudoku

Three Blocks Palindrome (hard version)

Robots on a Grid

Decimal

Forest Program

Angle Beats

Invoker

MUV LUV EXTRA

MUV LUV UNLIMITED

Escape

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