今天是毕业的学长给高二的同学测试。组队比赛,ACM赛制,于是就愉快的和学姐一队啦。
看到英文题面感到恐慌,不过好在不难读懂。
A:并没有什么技术含量的模拟题;
B:字符串题,给定一些比赛和每个队胜利的场数,判断是否合法。
[[a-b]-[c-d]],a 2,b 0,c,1,d 0就是合法的。(意思是:a和b比一场,c和d比一场,两场的胜者再一起比一场)
递归地读入形成一个树的结构,从最基础的比赛层层check上去。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 # include <cstdio> 2 # include <iostream> 3 # include <cstring> 4 # include <string> 5 6 using namespace std; 7 8 int l,a,b,v[30],x,cnt=26,ap=0; 9 bool vis[30],f=true; 10 char p,s[1000]; 11 struct race 12 { 13 int x,y,w; 14 }r[1000]; 15 16 int read (int x) 17 { 18 int pos=x+1,a,b; 19 if(s[pos]=='[') 20 { 21 pos=read(pos); 22 a=cnt; 23 }else 24 a=s[pos]-'a'; 25 pos+=2; 26 if(s[pos]=='[') 27 { 28 pos=read(pos); 29 b=cnt; 30 }else 31 b=s[pos]-'a'; 32 r[++cnt].x=a; 33 r[cnt].y=b; 34 return pos+1; 35 } 36 37 int main() 38 { 39 scanf("%s",s+1); 40 read(1); 41 l=strlen(s+1); 42 for (int i=1;i<=l;++i) 43 if(s[i]>='a'&&s[i]<='z'&&vis[ s[i]-'a' ]==false) 44 vis[ s[i]-'a' ]=true,ap++; 45 for (int i=1;i<=ap;++i) 46 { 47 cin>>p>>x; 48 v[p-'a']=x; 49 } 50 for (int i=27;i<=cnt;++i) 51 { 52 race now=r[i]; 53 if(now.x<26) 54 a=now.x; 55 else 56 a=r[now.x].w; 57 if(now.y<26) 58 b=now.y; 59 else 60 b=r[now.y].w; 61 if(v[a]==0&&v[b]!=0) 62 { 63 r[i].w=b; 64 v[b]--; 65 } 66 else if(v[a]!=0&&v[b]==0) 67 { 68 r[i].w=a; 69 v[a]--; 70 } 71 else f=false; 72 } 73 for (int i=0;i<26;++i) 74 if(v[i]) f=false; 75 if(f) printf("Yes "); 76 else printf("No "); 77 return 0; 78 }
C:求在$[l,r]$区间内素因子个数为质数的数。$1<=l,r<=1e9$
学姐做的这个题,是素数筛一类的思想。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <cmath> 3 4 const int N = 1000000; 5 6 bool notp[N+5]; 7 int prm[N+5], tot, a[N+5], b[N+5], l, r, ans; 8 9 void prime() { 10 notp[0] = notp[1] = 1; 11 for (int i = 2; i <= N; ++i) { 12 if (!notp[i]) prm[++tot] = i; 13 for (int j = 1; i * prm[j] <= N && j <= tot; ++j) { 14 notp[i*prm[j]] = 1; 15 if (i % prm[j] == 0) break; 16 } 17 } 18 } 19 20 int main() { 21 scanf("%d%d", &l, &r); 22 prime(); 23 for (int i = l; i <= r; ++i) b[i-l+1] = i; 24 for (int i = 1; i <= tot; ++i) { 25 if (prm[i] > r) break; 26 int j = l / prm[i]; 27 if (prm[i] * j < l) ++j; 28 for (j; prm[i] * j <= r; ++j) { 29 while (b[prm[i]*j-l+1] % prm[i] == 0) 30 ++a[prm[i]*j-l+1], b[prm[i]*j-l+1] /= prm[i]; 31 } 32 } 33 for (int i = l; i <= r; ++i) { 34 if (b[i-l+1] > 1) ++a[i-l+1]; 35 if (!notp[a[i-l+1]]) ++ans; 36 } 37 printf("%d ", ans); 38 return 0; 39 }
D,E,F,G:咕咕咕;
H:点开H题可能是我整场比赛最正确的决定了。正好是我之前想过的一个题,类似于那道L语言,但是要判断方案数,数据范围非常大,但是抱着反正不会交了不亏的思想写了一个交上去,A了...而且还比AC自动机快?果然做题要有信仰。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 # include <cstdio> 2 # include <iostream> 3 # include <cstring> 4 # include <string> 5 # define R register int 6 # define mod 1000000007 7 8 using namespace std; 9 10 const int maxn=200009; 11 int n,cnt=1; 12 char c[100005]; 13 int ch[maxn][26],l; 14 bool vis[maxn]; 15 int dp[100005]; 16 17 inline void ins() 18 { 19 int n,x=1,l=strlen(c+1); 20 for (R i=1;i<=l;++i) 21 { 22 n=c[i]-'a'; 23 if(!ch[x][n]) ch[x][n]=++cnt; 24 x=ch[x][n]; 25 } 26 vis[x]=true; 27 } 28 29 inline void Find (int pos) 30 { 31 int x=1,n; 32 for (R i=pos;i<=l;++i) 33 { 34 n=c[i]-'a'; 35 if(!ch[x][n]) break; 36 x=ch[x][n]; 37 if(vis[x]) 38 { 39 dp[i]+=dp[pos-1]; 40 if(dp[i]>=mod) dp[i]-=mod; 41 } 42 } 43 } 44 45 int main() 46 { 47 scanf("%d",&n); 48 for (R i=1;i<=n;++i) 49 { 50 scanf("%s",c+1); 51 ins(); 52 } 53 dp[0]=1; 54 scanf("%s",c+1); 55 l=strlen(c+1); 56 for (R i=0;i<l;++i) 57 if(dp[i]) Find(i+1); 58 printf("%d ",dp[l]); 59 return 0; 60 }
I,J,K:咕咕咕;
---shzr