I have long heard that NOI2015 questions are the easiest set of NOI questions. Even so, I still do super spicy chicken.
The program automatically analyzes this 'popular-' sb and check the set of questions, and I have nothing to say about the 'improved-' tree profile template question of the package manager .
The first question I encountered, which may be difficult to select, was the D1T3 sushi dinner.
The first reaction I saw was the model related to the bipartite graph in the network flow, but it has nothing to do with the bipartite graph.
I think it must have something to do with the root sign, but I rarely do such a question and I can't come up with any method.
Then give up treatment and fight violence, because prime numbers > n/2 can only appear in one number, so these prime numbers can be ignored, and only prime numbers with shape pressure <= n/2 can be used, there are 15 within 100, Then the 3-digit form pressed a violence, can you deceive 10 points more than 30 points of violence?
The positive solution is indeed related to the root sign. We consider that there are only 8 prime numbers <= sqrt(n), and each number contains at most one prime number > sqrt(n).
For numbers that do not contain primes > sqrt(n), we directly press dp, and for numbers containing primes > sqrt(n), we group them according to the number of primes, and press dp.
In the dp of each group, g[0/1][i][j] represents the number of schemes selected by the 0/1th person in this group.
The initial value of each group g[0][i][j]=g[1][i][j]=f[i][j]
The final answer f[i][j]=g[0][i][j]+g[1][i][j]-f[i][j] (minus this group of two people who have nothing not selected)
There seem to be a lot of questions that use the feature of "there is only one prime number containing > sqrt(n)", but I have written too few questions.
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=500+7,maxt=(1<<8)+7;
const ll prime[maxn]={2,3,5,7,11,13,17,19},totp=7,sz=(1<<8)-1;
ll n,mod,f[2][maxt][maxt],g[2][maxt][maxt],ans;
char cc; ll ff;
template<typename T>void read(T& aa) {
aa=0;cc=getchar();ff=1;
while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
if(cc=='-') ff=-1,cc=getchar();
while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
aa*=ff;
}
bool is_ok(int x) {
For(i,0,totp) while(x%prime[i]==0) x/=prime[i];
return x==1;
}
bool is_prime(int x) {
for(int i=2;i*i<=x;++i) if(x%i==0) return 0;
return 1;
}
void mo(ll& x) {if(x>=mod) x-=mod;}
int get_num(int x) {
int rs=0;
For(i,0,totp) if(x%prime[i]==0) rs|=(1<<i);
return rs;
}
int main() {
read(n); read(mod);
int o=0,x,up,r; f[0][0][0]=1;
For(i,2,n) if(is_ok(i)){
o^=1; x=get_num(i);
// For(s,0,sz) For(t,0,sz) f[o][s][t]=0;
memset(f[o],0,sizeof(f[o]));
For(s,0,sz) For(t,0,sz) if(f[o^1][s][t]) {
f[o^1][s][t]%=mod;
f[o][s][t]+=f[o^1][s][t];
if(!(x&t)) f[o][s|x][t]+=f[o^1][s][t];
if(!(x&s)) f[o][s][t|x]+=f[o^1][s][t];
}
}
For(i,23,n) if(is_prime(i)) {
o^=1; up=n/i;
For(s,0,sz) For(t,0,sz) f[o][s][t]=mod-(f[o^1][s][t]%=mod);
For(s,0,sz) For(t,0,sz) g[0][s][t]=g[1][s][t]=f[o^1][s][t];
For(j,1,up) {
x=get_num(j);
Rep(s,sz,0) Rep(t,sz,0) if(!(x&t)) {
r=g[0][s][t];
g [0] [s | x] [t] + = r; mo (g [0] [s | x] [t]);
}
Rep(s,sz,0) Rep(t,sz,0) if(!(x&s)) {
r=g[1][s][t];
g [1] [s] [t | x] + = r; mo (g [1] [s] [t | x]);
}
}
For(s,0,sz) For(t,0,sz)
f[o][s][t]=(f[o][s][t]+g[0][s][t]+g[1][s][t])%mod;
}
For(s,0,sz) For(t,0,sz) ans+=f[o][s][t];
printf("%lld\n",ans%mod);
return 0;
}
The second question I encountered that may have provincial difficulty is D2T1 Homer Epic
Seeing this question and thinking that I might be able to make it out, I feel like it's actually quite to my taste ( if I remember the Haverman tree )
And it's not that I haven't done this kind of "combined" type of questions, maybe I have A few more?
Later, I wrote a dp like O(n^2log(n)), and I expected 45 to write down and only 30.
I first think of this question as selecting n positions on the trie tree, and there is no ancestor-grandchild relationship between these n positions, so the depth is the length.
Because we know that if n positions are determined, then the depth with more occurrences must be small.
Then we consider the depth to increase from small to large. dp[i][j] represents the minimum cost of processing to the point with the largest number of occurrences at present, and there are j positions in this layer . len[i][j] represents processing to the point with the largest number of occurrences. A layer has j positions to choose the minimum depth at minimum cost.
Initially dp[0][k]=w[1], len[0][k]=1
When calculating dp[i], then enumerate how much the depth of this point is more than the depth of the previous point. If the number of optional positions is >=ni, you don't need to go further.
The positive solution is the k-forked Huffman tree. Note that the number should be supplemented to a multiple of (k-1) at the beginning.
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=1e5+7;
ll n,k,ans;
char cc; ll ff;
template<typename T>void read(T& aa) {
aa=0;cc=getchar();ff=1;
while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
if(cc=='-') ff=-1,cc=getchar();
while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
aa*=ff;
}
struct Node{
ll dep,w;
Node(){}
Node(ll dep,ll w):dep(dep),w(w){}
bool operator < (const Node& b) const{return w==b.w? dep>b.dep:w>b.w;}
}O;
priority_queue<Node> G;
int main() {
read(n); read(k); ll x,d;
For(i,1,n) {
read(x);
G.push(Node(0,x));
}
if((n-1)%(k-1)!=0) Rep(i,k-1-(n-1)%(k-1),1) G.push(Node(0,0));
while(G.size()>1) {
x=0; d=0;
For(i,1,k) {
o=G.top(); G.pop();
x+=o.w; d=max(d,o.dep);
ans+=ow;
}
G.push(Node(d+1,x));
}
o = G.top ();
printf("%lld\n%lld\n",ans,o.dep);
return 0;
}
As for the wine tasting...
I have done it twice before, this is the 3rd time, here is a link: http://www.cnblogs.com/Serene-shixinyi/p/8321281.html