Blog garden hung up, not sticky map.
Write simple point.
#1：100+100+25=225
#2：100+70+35=205
# 2: 60 + 100 + 45 = 205 (I)
Back to the first room pretty good first battle.
After I finished I thought the AK card however is often labeled as violent T2, T3 of perjury greedy (although a game-high points)
The whole thinking. very good.
Maintained.
Note that constant, constant questions on the card you want to take the time to optimize the play card often.
T1：Simple
It is a more sucker.
Under prime to go do, so to get rid of all nm gcd, the answer to this first contribution count.
We consider listing a table, each row of n (n <= m).
If this number is a multiple of m, from here then this column does not appear the number of bad.
So long as each column is considered a multiple of the earliest when it will appear m just fine, ex_gcd.
1 #include<cstdio> 2 #include<iostream> 3 using namespace std; 4 #define int long long 5 int gcd(int a,int b){return b?gcd(b,a%b):a;} 6 void ex_gcd(int a,int b,int &x,int &y){ 7 if(!b){x=1;y=0;return;} 8 ex_gcd(b,a%b,x,y); 9 int r=x;x=y;y=r-a/b*y; 10 } 11 main(){//freopen("ex_simple2.in","r",stdin); 12 int t;scanf("%lld",&t); 13 while(t--){ 14 int n,m,q,ans=0,x,y,g; 15 scanf("%lld%lld%lld",&n,&m,&q); 16 if(n>m)n^=m^=n^=m; 17 g=gcd(n,m);ans+=q-q/g; 18 n/=g;m/=g;q/=g; 19 ex_gcd(n,m,x,y);x%=m;//printf("x=%lld\n",x); 20 for(int i=1;i<n;++i)ans+=min(((m-i)*x%m+m)%m,(q-i+n)/n);//,printf("%lld\n",ans); 21 printf("%lld\n",ans); 22 } 23 }
T2：Walk
Often title card.
W of less than 1 million up to about 240 number, so do the longest chain dp find each divisor like.
1 // Each number is within 240 to about 1,000,000 have the largest number, and an average of only 14 2 // Note card memory: 61079552 int . 3 #include <cstdio> . 4 #include <the iostream> . 5 #include <unordered_map> . 6 the using namespace STD; . 7 int the FIR [ 1000005 ], L [ 14000000 ], the TO [ 14000000 ], the CNT; . 8 void the Add ( int A, int D) {L [the CNT ++] = the FIR [A]; the FIR [A] = the CNT; the tO [the CNT] = D;} . 9 int FIR [ 400005 ], L [ 800 005 ], to [ 800 005 ], W [800005],cnt; 10 void link(int a,int b,int v){l[++cnt]=fir[a];fir[a]=cnt;to[cnt]=b;w[cnt]=v;} 11 int ans[400005],n; 12 unordered_map<int,int>dp[400005]; 13 int read(){ 14 register int p=0;register char ch=getchar(); 15 while(ch<'0'||ch>'9')ch=getchar(); 16 while(ch>='0'&&ch<='9')p=(p<<3)+(p<<1)+ch-48,ch=getchar(); 17 return p; 18 } 19 void dfs(int p,int fa){ 20 for(int i=fir[p];i;i=l[i])if(to[i]!=fa){ 21 dfs(to[i],p); 22 for(int j=FIR[w[i]];j;j=L[j]){ 23 int x=dp[to[i]][TO[j]],y=dp[p][TO[j]]; 24 ans[x+y+1]=max(ans[x+y+1],TO[j]); 25 dp[p][TO[j]]=max(y,x+1); 26 } 27 dp[to[i]].clear(); 28 } 29 } 30 int main(){//freopen("t2.in","r",stdin);freopen("my.out","w",stdout); 31 n=read();int mx=0; 32 for(int x,y,w,i=1;i<n;++i)x=read(),y=read(),w=read(),link(x,y,w),link(y,x,w),mx=max(mx,w); 33 for(int i=1;i<=mx;++i)for(int j=i;j<=mx;j+=i)add(j,i); 34 dfs(1,0); 35 for(int i=n;i;--i)ans[i]=max(ans[i],ans[i+1]); 36 for(int i=1;i<=n;++i)printf("%d\n",ans[i]); 37 }
Then the T violent.
For each construction to about several sides, rather than root hard sieve pretreatment on the A.
1 #include<cstdio> 2 #include<iostream> 3 #include<vector> 4 using namespace std; 5 vector<int>a[1000004],b[1000005]; 6 int fir[400005],l[800005],to[800005],cnt; 7 void link(int a,int b){l[++cnt]=fir[a];fir[a]=cnt;to[cnt]=b;} 8 int ans[400005],lgst; 9 int read(){ 10 register int p=0;register char ch=getchar(); 11 while(ch<'0'||ch>'9')ch=getchar(); 12 while(ch>='0'&&ch<='9')p=(p<<3)+(p<<1)+ch-48,ch=getchar(); 13 return p; 14 } 15 int dfs(int p,int fa){ 16 int mx=0; 17 for(int i=fir[p];i;i=l[i])if(to[i]!=fa){ 18 int x=dfs(to[i],p)+1; 19 lgst=max(lgst,mx+x);mx=max(mx,x); 20 }fir[p]=0;return mx; 21 } 22 int main(){//freopen("ex_walk2.in","r",stdin);freopen("my.out","w",stdout); 23 register int n=read(); 24 for(int t=1,x,y,w;t<n;++t){ 25 x=read(),y=read(),w=read(); 26 for(int i=1;i*i<=w;++i)if(w%i==0){ 27 a[i].push_back(x),b[i].push_back(y); 28 if(i*i!=w)a[w/i].push_back(x),b[w/i].push_back(y); 29 } 30 } 31 for(int i=1;i<=1000000;++i){ 32 for(int j=0;j<a[i].size();++j)link(a[i][j],b[i][j]),link(b[i][j],a[i][j]); 33 for(int j=0;j<a[i].size();++j)if(fir[a[i][j]])dfs(a[i][j],0); 34 ans[lgst]=max(ans[lgst],i); 35 cnt=lgst=0; 36 } 37 for(int i=n;i;--i)ans[i]=max(ans[i],ans[i+1]); 38 for(int i=1;i<=n;++i)printf("%d\n",ans[i]); 39 }
T3：Travel
I understand the thinking, but also think of optimization.
But afternoon will not have time to test apparently changed.