Codeforces Round #538 (Div. 2)
Codeforces 1114
貌似最近平均难度最低的一场div2了...
但我没有把握住机会TAT
D题没往DP想 写模拟自闭了40多分钟...才想起是个...最简单的区间DP
再多二十分钟就能调出F的傻逼错误了...
A.Got Any Grapes?
puts
的返回值原来是0。。刚开始写的return !puts("NO");
在样例上RE了一次= =
#include <set>
#include <map>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cstring>
#include <algorithm>
#define pc putchar
#define gc() getchar()
typedef long long LL;
inline int read()
{
int now=0,f=1;register char c=gc();
for(;!isdigit(c);c=='-'&&(f=-1),c=gc());
for(;isdigit(c);now=now*10+c-48,c=gc());
return now*f;
}
int main()
{
int x=read(),y=read(),z=read(),a=read(),b=read(),c=read();
if(a<x) return puts("NO"),0;
a-=x;
if(a+b<y) return puts("NO"),0;
int tot=a+b+c-y-z;
if(tot<0) return puts("NO"),0;
return puts("YES"),0;
return 0;
}
B.Yet Another Array Partitioning Task
因为可以每隔\(m\)个分一段,所以选出最大的\(m*k\)个数,每隔\(m\)个分一段就行了。
#include <set>
#include <map>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cstring>
#include <algorithm>
#define pc putchar
#define gc() getchar()
typedef long long LL;
const int N=2e5+5;
struct Node
{
int v,p;
bool operator <(const Node &x)const
{
return v>x.v;
}
}A[N];
inline int read()
{
int now=0,f=1;register char c=gc();
for(;!isdigit(c);c=='-'&&(f=-1),c=gc());
for(;isdigit(c);now=now*10+c-48,c=gc());
return now*f;
}
int main()
{
static int Ans[N];
static bool vis[N];
int n=read(),m=read(),K=read();
for(int i=1; i<=n; ++i) A[i]=(Node){read(),i};
std::sort(A+1,A+1+n);
LL sum=0;
for(int i=1; i<=K*m; ++i) vis[A[i].p]=1, sum+=A[i].v;
int cnt=0;
for(int i=1,t=0; i<=n; ++i)
if(vis[i] && ++t==m)
t=0, Ans[++cnt]=i;
printf("%I64d\n",sum);
for(int i=1; i<cnt; ++i) printf("%d ",Ans[i]);
return 0;
}
C.Trailing Loves (or L'oeufs?)
\(Description\)
给定\(n,b\),求\(b\)进制下\(n!\)的末尾零的个数。
\(n\leq10^{18},\ b\leq10^{12}\)。
\(Solution\)
经典问题。。所以也是一道能搜到的题。。
如果\(d=10\),就是求\(n!\)有多少个\(2\)和\(5\)的因子\(c_2,c_5\),答案是\(\min\{c_2,c_5\}\),不难理解。
如果\(d\neq10\),同样对\(d\)质因数分解,令\(d=\prod_{i=1}^kp_i^{a_i},\ a_i\neq0\),依次求出\(n!\)中有多少个\(p_i\),记为\(c_i\),答案就是\(\min_{i=1}^k\{c_i\}\)。
求\(n!\)中质因子\(p\)的个数的公式:$$f(n)=\left\lfloor\frac{n}{p}\right\rfloor +\left\lfloor\frac{n}{p^2}\right\rfloor +\left\lfloor\frac{n}{p^3}\right\rfloor +\cdots$$
实际写的时候\(n\)每次除以\(p\)即可,不需要分母不断乘\(p\),会爆long long
。。。
#include <set>
#include <map>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cstring>
#include <algorithm>
#define pc putchar
#define gc() getchar()
typedef long long LL;
const int N=1e6+5;
inline LL read()
{
LL now=0;register char c=gc();
for(;!isdigit(c);c=gc());
for(;isdigit(c);now=now*10+c-48,c=gc());
return now;
}
LL Calc(LL x,LL y)
{
LL res=0;
for(; x; x/=y) res+=x/y;
return res;
}
int main()
{
static LL P[N];
static int tm[N];
LL n=read(),b=read();
//1
LL ans=1ll<<60;
for(int i=2; 1ll*i*i<=b; ++i)
if(!(b%i))
{
LL cnt=1; b/=i;
while(!(b%i)) b/=i, ++cnt;
ans=std::min(ans,Calc(n,i)/cnt);
}
if(b!=1) ans=std::min(ans,Calc(n,b));
printf("%I64d\n",ans);
return 0;
//2
int t=0;
for(int i=2; 1ll*i*i<=b; ++i)
if(!(b%i))
{
P[++t]=i, b/=i, tm[t]=1;
while(!(b%i)) b/=i, ++tm[t];
}
if(b!=1) P[++t]=b, tm[t]=1;
ans=1ll<<60;
for(int i=1; i<=t; ++i)
{
LL cnt=0;
// for(LL x=P[i]; x<=n&&x>0; x*=P[i]) cnt+=n/x;//这么写可能直接爆longlong爆成正数啊。。。= =
for(LL x=n; x; x/=P[i]) cnt+=x/P[i];
ans=std::min(ans,cnt/tm[i]);
}
printf("%I64d\n",ans);
return 0;
}
D.Flood Fill(区间DP)
刚开始一看\(O(n^2)\),就想枚举起点然后模拟。。(mdzz)
无脑区间DP。。
\(f[i][j][0/1]\)表示合并完\(i\sim j\)区间,现在颜色是\(c_i/c_j\)的最小花费。转移特判一下即可。
或者先去掉相邻的重复元素,直接\(f[i][j]\)表示合并完\(i\sim j\)区间的最小花费。如果\(c_i=c_j\),\(f[i][j]=f[i+1][j-1]+1\);否则\(f[i][j]=\min(f[i+1][j],\ f[i][j-1])+1\)。
#include <set>
#include <map>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cstring>
#include <algorithm>
#define pc putchar
#define gc() getchar()
typedef long long LL;
const int N=5005;
int A[N],f[N][N][2];
inline int read()
{
int now=0,f=1;register char c=gc();
for(;!isdigit(c);c=='-'&&(f=-1),c=gc());
for(;isdigit(c);now=now*10+c-48,c=gc());
return now*f;
}
bool Check(int n)
{
for(int i=1; i<=n; ++i) if(A[i]!=A[1]) return 0;
puts("0");
return 1;
}
int main()
{
int n=read();
for(int i=1; i<=n; ++i) A[i]=read();
if(Check(n)) return 0;
memset(f,0x3f,sizeof f);
for(int i=1; i<=n; ++i) f[i][i][0]=f[i][i][1]=0;
for(int l=1; l<n; ++l)
for(int i=1; i+l<=n; ++i)
{
int j=i+l;
f[i][j][0]=std::min(f[i][j-1][0]+2-2*(A[i]==A[j]),std::min(f[i][j-1][1]+2-(A[j-1]==A[j])-(A[i]==A[j]),std::min(f[i+1][j][0]+1-(A[i]==A[i+1]),f[i+1][j][1]+1-(A[j]==A[i]))));
f[i][j][1]=std::min(f[i][j-1][0]+1-(A[i]==A[j]),std::min(f[i][j-1][1]+1-(A[j-1]==A[j]),std::min(f[i+1][j][0]+2-(A[i]==A[i+1])-(A[i]==A[j]),f[i+1][j][1]+2-2*(A[j]==A[i]))));
}
printf("%d\n",std::min(f[1][n][0],f[1][n][1]));
return 0;
}
E.Arithmetic Progression(交互 二分 随机化)
首先通过操作二可以二分出最大值。
然后剩下\(30\)次操作一能干什么呢。。只能帮我们确定出数列中有某些数,那能做的好像就是随机确定\(30\)个数。
等差数列中任意两个数作差可以得到\(x\cdot d\),把确定出的数两两作差然后对差求\(\gcd\),是有可能得到真正的\(d\)的。
正确概率是多少呢,感觉挺高的。。然后有最大值有公差,就做完了。
官方题解中对正确概率有证明,大约是\(1.86185\times10^{-9}\)。
做差的时候先排序,只求出相邻两个数的差就够了,并不需要所有数两两之间作差。
#include <ctime>
#include <cstdio>
#include <cctype>
#include <algorithm>
#define gc() getchar()
#define Flush() fflush(stdout)
typedef long long LL;
const int N=1e6+5;
int id[N];
inline int read()
{
int now=0;register char c=gc();
for(;!isdigit(c);c=gc());
for(;isdigit(c);now=now*10+c-48,c=gc());
return now;
}
inline int Rd()
{
return rand()<<15|rand();
}
inline int Query1(int n)
{
int p=Rd()%n+1;
printf("? %d\n",id[p]), Flush();
std::swap(id[p],id[n]);//不这么写也行 差别不大 无所谓啦
return read();
}
inline int Query2(int x)
{
printf("> %d\n",x), Flush();
return read();
}
int main()
{
static int A[66];
srand(time(0));
int n=read();
int l=0,r=1e9,mid,mx=0,rest=60;
while(l<=r)
if(--rest,Query2(mid=l+r>>1)) mx=l=mid+1;
else r=mid-1;
for(int i=1; i<=n; ++i) id[i]=i;
int t=0; rest=std::min(rest,n);
for(int now=n; rest; --rest) A[++t]=Query1(now--);
std::sort(A+1,A+1+t), A[++t]=mx;
int d=A[2]-A[1];
for(int i=3; i<=t; ++i) if(A[i]!=A[i-1]) d=std::__gcd(d,A[i]-A[i-1]);
printf("! %d %d\n",mx-(n-1)*d,d), Flush();
return 0;
}
F.Please, another Queries on Array?(线段树 欧拉函数)
\(Description\)
给定长为\(n\)的序列\(A_i\),要求支持两种操作:
- \(l,r,v\),区间\([l,r]\)乘一个数\(v\);
- \(l,r\),输出\(\varphi(\prod_{i=l}^rA_i)\)。
\(n\leq4\times10^5,\ q\leq2\times10^5,\ A_i,v\leq300\)。
\(Solution\)
听旁边dalao讨论好像特别可做,扔掉E看F。
忘了欧拉函数怎么求。。去百度百科看了下,发现这不是线段树裸题吗。。
设\(\prod_{i=l}^rA_i=n=\prod_{i=1}^kp_i^{a_i}\),那\(\varphi(n)=\prod_{i=1}^kp_i^{a_i}\frac{p_i-1}{p_i}=n\prod_{i=1}^k\frac{p_i-1}{p_i}\)。
\(k\)是\(300\)以内质数个数,就是\(62\)。所以我们维护一个区间乘积、区间某个质因子是否出现过就完了。
后者可以用bitset
维护,或者因为只有\(62\)个用long long
也行。
复杂度\(O(q(\log^2n+62))\)(区间乘积竟然第一次写...要快速幂所以多个\(\log\)。但是有方法可以去掉快速幂的复杂度,见这一题)。
//1809ms 42100KB
#include <cstdio>
#include <cctype>
#include <bitset>
#include <cstring>
#include <algorithm>
#define gc() getchar()
#define mod 1000000007
typedef long long LL;
const int N=4e5+5,M=301,LIM=62;
const int P[LIM]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293};
const int ref[M]={0,0,0,1,0,2,0,3,0,0,0,4,0,5,0,0,0,6,0,7,0,0,0,8,0,0,0,0,0,9,0,10,0,0,0,0,0,11,0,0,0,12,0,13,0,0,0,14,0,0,0,0,0,15,0,0,0,0,0,16,0,17,0,0,0,0,0,18,0,0,0,19,0,20,0,0,0,0,0,21,0,0,0,22,0,0,0,0,0,23,0,0,0,0,0,0,0,24,0,0,0,25,0,26,0,0,0,27,0,28,0,0,0,29,0,0,0,0,0,0,0,0,0,0,0,0,0,30,0,0,0,31,0,0,0,0,0,32,0,33,0,0,0,0,0,0,0,0,0,34,0,35,0,0,0,0,0,36,0,0,0,0,0,37,0,0,0,38,0,0,0,0,0,39,0,0,0,0,0,40,0,41,0,0,0,0,0,0,0,0,0,42,0,43,0,0,0,44,0,45,0,0,0,0,0,0,0,0,0,0,0,46,0,0,0,0,0,0,0,0,0,0,0,47,0,0,0,48,0,49,0,0,0,50,0,0,0,0,0,51,0,52,0,0,0,0,0,0,0,0,0,53,0,0,0,0,0,54,0,0,0,0,0,55,0,0,0,0,0,56,0,57,0,0,0,0,0,58,0,0,0,59,0,60,0,0,0,0,0,0,0,0,0,61};
//为什么const之后慢了不少= =
inline int read();
struct Segment_Tree
{
#define ls rt<<1
#define rs rt<<1|1
#define lson l,m,ls
#define rson m+1,r,rs
#define S N<<2
int val[S],tag[S];
LL now,Ans,f[S],tagf[S];
#undef S
#define Upd(rt,v,l,vf) val[rt]=1ll*val[rt]*FP(v,l)%mod, tag[rt]=1ll*tag[rt]*v%mod, f[rt]|=vf, tagf[rt]|=vf
#define Update(rt) f[rt]=f[ls]|f[rs], val[rt]=1ll*val[ls]*val[rs]%mod
inline int FP(int x,int k)
{
int t=1;
for(; k; k>>=1,x=1ll*x*x%mod)
if(k&1) t=1ll*t*x%mod;
return t;
}
void Build(int l,int r,int rt)
{
tag[rt]=1;
if(l==r)
{
int v=read(); val[rt]=v;
for(int i=0; i<LIM&&P[i]*P[i]<=v; ++i)
if(!(v%P[i]))
{
f[rt]|=1ll<<i, v/=P[i];//1ll!!!
while(!(v%P[i])) v/=P[i];
}
if(v!=1) f[rt]|=1ll<<ref[v];
return;
}
int m=l+r>>1;
Build(lson), Build(rson), Update(rt);
}
inline void PushDown(int rt,int m)
{
int l=ls,r=rs;
Upd(l,tag[rt],m-(m>>1),tagf[rt]), Upd(r,tag[rt],m>>1,tagf[rt]), tag[rt]=1, tagf[rt]=0;
}
void Modify(int l,int r,int rt,int L,int R,int v)
{
if(L<=l && r<=R)
{
Upd(rt,v,r-l+1,now);
return;
}
if(tag[rt]!=1) PushDown(rt,r-l+1);
int m=l+r>>1;
if(L<=m) Modify(lson,L,R,v);
if(m<R) Modify(rson,L,R,v);
Update(rt);
}
int Query(int l,int r,int rt,int L,int R)
{
if(L<=l && r<=R) return Ans|=f[rt],val[rt];
if(tag[rt]!=1) PushDown(rt,r-l+1);
int m=l+r>>1;
if(L<=m)
if(m<R) return 1ll*Query(lson,L,R)*Query(rson,L,R)%mod;
else return Query(lson,L,R);
return Query(rson,L,R);
}
}T;
inline int read()
{
int now=0;register char c=gc();
for(;!isdigit(c);c=gc());
for(;isdigit(c);now=now*10+c-48,c=gc());
return now;
}
//void Init(int n)
//{
// static int P[N];
// static bool notP[M];
// for(int i=2,cnt=0; i<=n; ++i)
// {
// if(!notP[i]) ref[i]=cnt, P[cnt++]=i;
// for(int j=0; j<cnt && i*P[j]<=n; ++j)
// {
// notP[i*P[j]]=1;
// if(!(i%P[j])) break;
// }
// }
//}
inline char GetOpt()
{
register char c;
while(!isalpha(c=gc()));
return c;
}
int main()
{
static int inv[N],coef[N];
// Init(M-1);
inv[1]=1;
for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
int n=read(),q=read();
T.Build(1,n,1);
while(q--)
{
if(GetOpt()=='T')
{
T.Ans=0;
int l=read(),r=read(),val=T.Query(1,n,1,l,r);
for(int i=0; i<LIM; ++i)
if(T.Ans>>i&1) val=1ll*val*coef[i]%mod;
printf("%d\n",val);
}
else
{
int l=read(),r=read(),x=read(),v=x;
LL now=0;
for(int i=0; i<LIM&&P[i]*P[i]<=v; ++i)
if(!(v%P[i]))
{
now|=1ll<<i, v/=P[i];
while(!(v%P[i])) v/=P[i];
}
if(v!=1) now|=1ll<<ref[v];
T.now=now, T.Modify(1,n,1,l,r,x);
}
}
return 0;
}
很久以前的奇怪但现在依旧成立的签名
attack is our red sun $$\color{red}{\boxed{\color{red}{attack\ is\ our\ red\ sun}}}$$ ------------------------------------------------------------------------------------------------------------------------