https://www.nowcoder.com/acm/contest/200#question
最近突然找到了打比赛的乐趣,于是参加了这场比赛.
生日宴会:https://www.nowcoder.com/acm/contest/200/A
按照题意模拟,没什么好说的.
1 # include <cstdio> 2 # include <iostream> 3 # include <cstring> 4 # include <cmath> 5 # include <algorithm> 6 # include <string> 7 # define R register int 8 9 using namespace std; 10 11 const int maxn=1005; 12 char name[maxn][13]; 13 int n,m,dat,x,y; 14 struct nod 15 { 16 int val,k; 17 }id[1240][maxn]; 18 int h[1240]; 19 20 bool cmp (nod a,nod b) 21 { 22 return a.val<b.val; 23 } 24 25 int main() 26 { 27 scanf("%d%d",&n,&m); 28 for (R i=1;i<=n;++i) 29 { 30 scanf("%s",name[i]); 31 scanf("%d",&dat); 32 id[dat%10000][ ++h[dat%10000] ].k=i; 33 id[dat%10000][ h[dat%10000] ].val=dat/10000; 34 } 35 for (R i=0;i<=1231;++i) 36 sort(id[i]+1,id[i]+1+h[i],cmp); 37 for (R i=1;i<=m;++i) 38 { 39 scanf("%d%d",&x,&y); 40 printf("%s ",name[ id[y][x].k ]); 41 } 42 return 0; 43 }
数据结构:https://www.nowcoder.com/acm/contest/200/B
题意概述:维护一个区间,支持区间加,区间乘,查询区间和,区间平方和.
好麻烦的题目啊,本来要写线段树的,后来觉得分块可能简单点就写了分块.事实上...并没有简单,还是要打很多标记,维护起来差不多麻烦.
1 # include <cstdio> 2 # include <iostream> 3 # include <cstring> 4 # include <cmath> 5 # include <algorithm> 6 # include <string> 7 # define R register int 8 # define ll long long 9 10 using namespace std; 11 12 const int maxn=10004; 13 long long x,a[maxn],s[maxn],ss[maxn],delta_a[maxn],delta_b[maxn]; 14 int b[maxn],n,m,siz,l,r,opt,si[maxn]; 15 16 ll read() 17 { 18 long long x=0,f=1; 19 char c=getchar(); 20 while (!isdigit(c)) { if(c=='-') f=-f; c=getchar(); } 21 while (isdigit(c)) { x=(x<<3)+(x<<1)+(c^48); c=getchar(); } 22 return x*f; 23 } 24 25 void down (int x) 26 { 27 s[x]=0; 28 ss[x]=0; 29 for (R i=1+siz*(x-1);i<=min(x*siz,n);++i) 30 { 31 a[i]=a[i]*delta_a[x]+delta_b[x]; 32 s[x]+=a[i]; 33 ss[x]+=a[i]*a[i]; 34 } 35 delta_a[x]=1; 36 delta_b[x]=0; 37 } 38 39 long long ask1 (int l,int r) 40 { 41 long long ans=0; 42 int id; 43 if(b[l]==b[r]) 44 { 45 id=b[l]; 46 down(id); 47 for (R i=l;i<=r;++i) ans+=a[i]; 48 } 49 else 50 { 51 id=b[l]; 52 down(id); 53 for (R i=l;i<=siz*b[l];++i) ans+=a[i]; 54 for (R i=b[l]+1;i<=b[r]-1;++i) ans+=s[i]; 55 id=b[l]; 56 down(id); 57 for (R i=1+siz*(b[r]-1);i<=r;++i) ans+=a[i]; 58 } 59 return ans; 60 } 61 62 long long ask2 (int l,int r) 63 { 64 int id; 65 long long ans=0; 66 if(b[l]==b[r]) 67 { 68 id=b[l]; 69 down(id); 70 for (R i=l;i<=r;++i) ans+=a[i]*a[i]; 71 } 72 else 73 { 74 id=b[l]; 75 down(id); 76 for (R i=l;i<=siz*b[l];++i) ans+=a[i]*a[i]; 77 for (R i=b[l]+1;i<=b[r]-1;++i) ans+=ss[i]; 78 id=b[l]; 79 down(id); 80 for (R i=1+siz*(b[r]-1);i<=r;++i) ans+=a[i]*a[i]; 81 } 82 return ans; 83 } 84 85 void add (int l,int r,ll c) 86 { 87 int id; 88 if(b[l]==b[r]) 89 { 90 id=b[l]; 91 down(id); 92 for (R i=l;i<=r;++i) 93 { 94 s[id]+=c; 95 ss[id]+=c*c+2*c*a[i]; 96 a[i]=a[i]+c; 97 } 98 } 99 else 100 { 101 id=b[l]; 102 down(id); 103 for (R i=l;i<=siz*b[l];++i) 104 { 105 s[id]+=c; 106 ss[id]+=c*c+2*c+a[i]; 107 a[i]=a[i]+c; 108 } 109 for (R i=b[l]+1;i<=b[r]-1;++i) 110 { 111 delta_a[i]+=c; 112 ss[i]+=c*c*si[i]+2*c*s[i]; 113 s[i]+=si[i]*c; 114 } 115 id=b[r]; 116 down(id); 117 for (R i=1+siz*(b[r]-1);i<=r;++i) 118 { 119 s[id]+=c; 120 ss[id]+=c*c+2*c+a[i]; 121 a[i]=a[i]+c; 122 } 123 } 124 } 125 126 void mul (int l,int r,ll c) 127 { 128 int id; 129 if(b[l]==b[r]) 130 { 131 id=b[l]; 132 down(id); 133 for (R i=l;i<=r;++i) 134 { 135 s[id]+=a[i]*(c-1); 136 ss[id]+=(c*c-1)*(a[i]*a[i]); 137 a[i]*=c; 138 } 139 } 140 else 141 { 142 id=b[l]; 143 down(id); 144 for (R i=l;i<=siz*b[l];++i) 145 { 146 s[id]+=a[i]*(c-1); 147 ss[id]+=(c*c-1)*(a[i]*a[i]); 148 a[i]*=c; 149 } 150 for (R i=b[l]+1;i<=b[r]-1;++i) 151 { 152 delta_b[i]*=c; 153 delta_a[i]*=c; 154 ss[i]*=c; 155 s[i]*=c; 156 } 157 id=b[r]; 158 down(id); 159 for (R i=1+siz*(b[r]-1);i<=r;++i) 160 { 161 s[id]+=a[i]*(c-1); 162 ss[id]+=(c*c-1)*(a[i]*a[i]); 163 a[i]*=c; 164 } 165 } 166 } 167 168 int main() 169 { 170 scanf("%d%d",&n,&m); 171 for (R i=1;i<=n;++i) a[i]=read(); 172 siz=sqrt(n); 173 for (R i=1;i<=n;++i) b[i]=(i-1)/siz+1; 174 for (R i=1;i<=n;++i) s[ b[i] ]+=a[i],ss[ b[i] ]+=a[i]*a[i]; 175 for (R i=1;i<=n;++i) si[ b[i] ]=siz,delta_a[i]=1; 176 si[ b[n] ]=n-siz*(b[n]-1); 177 for (R i=1;i<=m;++i) 178 { 179 scanf("%d",&opt); 180 if(opt==1||opt==2) 181 { 182 scanf("%d%d",&l,&r); 183 if(opt==1) printf("%lld ",ask1(l,r)); 184 else printf("%lld ",ask2(l,r)); 185 } 186 else 187 { 188 scanf("%d%d%lld",&l,&r,&x); 189 if(opt==3) mul(l,r,x); 190 else add(l,r,x); 191 } 192 } 193 return 0; 194 }
迎风舞:https://www.nowcoder.com/acm/contest/200/E
题意概述:物理题.
三分抛出方向与水平方向的夹角,解方程算出时间和落地时间即可.关于这个函数为什么是单峰的,这里提供一个不严谨的证明:首先直觉上平抛肯定是比不上上抛的,但是角度为$90$度的上抛运动等于没有抛出去,所以函数应该是满足三分性的.
1 # include <cstdio> 2 # include <iostream> 3 # include <cstring> 4 # include <cmath> 5 # include <algorithm> 6 # include <string> 7 # define R register int 8 9 using namespace std; 10 11 const double g=9.80665; 12 const double eps=0.000001; 13 double h,v,l,r,lmid,rmid,ans,lans; 14 int T; 15 16 double f (double si) 17 { 18 double vx,vy,a,b,c,delta,x1,x2,t; 19 vy=v*si; 20 vx=sqrt(v*v-vy*vy); 21 a=-g/2.0; 22 b=vy; 23 c=h; 24 delta=sqrt(b*b-4.0*a*c); 25 26 x1=(-b+delta)/(2*a); 27 x2=(-b-delta)/(2*a); 28 29 t=max(x1,x2); 30 return t*vx; 31 } 32 33 int main() 34 { 35 scanf("%d",&T); 36 while (T--) 37 { 38 scanf("%lf%lf",&h,&v); 39 l=0,r=1; 40 while (r-l>eps) 41 { 42 lmid=l+(r-l)/3.0; 43 rmid=r-(r-l)/3.0; 44 if(f(lmid)<f(rmid)) 45 l=lmid; 46 else r=rmid; 47 } 48 printf("%.5lf ",f(l)); 49 } 50 return 0; 51 }
------------------比赛时就做到这里了,以下是赛后补题------------------
随风飘:https://www.nowcoder.com/acm/contest/200/D
题意概述:给定$n$个字符串,求它们两两之间的$LCP$的长度和,此外给出$T$组询问$k$,表示有$k$个字符串消失了,求所有的消失情况下$LCP$的长度和的和.
其实考试时本来是有机会$A$的,但是都以为开不下存字符串的数组导致纷纷弃疗.开数组靠的是信仰...其实当时为什么不开一个$vector$呢,果然考试时智商会降低.
首先把字符串都串到$Trie$树上去,统计$k=0$时的答案.对于其他的询问就不能再上树了!要考虑运用组合数学.每个$LCP$有$C_{n-2}^k$种方法留下来,直接算就好了.
1 # include <cstdio> 2 # include <iostream> 3 # include <cstring> 4 # include <cmath> 5 # include <algorithm> 6 # include <string> 7 # define R register int 8 # define ll long long 9 # define mod 1000000007 10 11 using namespace std; 12 13 const int maxs=3000006; 14 int k,n,q,len,cnt=1,c[4003][302]; 15 char s[maxs]; 16 17 struct Trie 18 { 19 int ch[maxs][26]; 20 int vis[maxs],s[maxs]; 21 long long ans; 22 void clear() 23 { 24 ans=0; 25 memset(ch,0,sizeof(ch)); 26 memset(vis,0,sizeof(vis)); 27 memset(s,0,sizeof(s)); 28 } 29 void ins (char *s) 30 { 31 int len=strlen(s); 32 int x=1; 33 for (R i=0;i<len;++i) 34 { 35 int c=s[i]-'a'; 36 if(!ch[x][c]) 37 ch[x][c]=++cnt; 38 x=ch[x][c]; 39 } 40 ans=(ans+vis[x]*len)%mod; 41 vis[x]++; 42 } 43 void upd (int x,int dep) 44 { 45 s[x]=vis[x]; 46 for (R i=0;i<26;++i) 47 if(ch[x][i]) 48 { 49 upd(ch[x][i],dep+1); 50 ans=(ans+1LL*dep*s[x]%mod*(s[ ch[x][i] ]))%mod; 51 s[x]+=s[ ch[x][i] ]; 52 } 53 } 54 }T; 55 56 int main() 57 { 58 scanf("%d%d",&n,&q); 59 for (R i=1;i<=n;++i) 60 { 61 scanf("%s",s); 62 T.ins(s); 63 } 64 T.upd(1,0); 65 c[0][0]=1; 66 for (R i=1;i<=n;++i) 67 { 68 c[i][0]=1; 69 for (R j=1;j<=min(i,300);++j) 70 c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod; 71 } 72 for (R i=1;i<=q;++i) 73 { 74 scanf("%d",&k); 75 printf("%lld ",1LL*c[n-2][k]*T.ans%mod); 76 } 77 return 0; 78 }