今天老师（orz sansirowaltz）让我们做了很久之前的一场Codeforces Round #257 (Div. 1)，这里给出A~C的题解，对应DIV2的C~E。
1 second
256 megabytes
standard input
standard output
Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times. Each cut must meet the following requirements:
- each cut should be straight (horizontal or vertical);
- each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
- each cut should go inside the whole chocolate bar, and all cuts must be distinct.
The picture below shows a possible way to cut a 5 × 6 chocolate for 5 times.
Imagine Jzzhu have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, Jzzhu wants this piece to be as large as possible. What is the maximum possible area of smallest piece he can get with exactly k cuts? The area of a chocolate piece is the number of unit squares in it.
A single line contains three integers n, m, k (1 ≤ n, m ≤ 109; 1 ≤ k ≤ 2·109).
Output a single integer representing the answer. If it is impossible to cut the big chocolate k times, print -1.
3 4 1
6
6 4 2
8
2 3 4
-1
In the first sample, Jzzhu can cut the chocolate following the picture below:
In the second sample the optimal division looks like this:
In the third sample, it's impossible to cut a 2 × 3 chocolate 4 times.
这题是数学问题，感觉没什么好说的，就是注意long long和细节，很容易WA掉。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std; int k;
long long n,m;
long long ans; int main()
{
scanf("%I64d%I64d%d",&n,&m,&k);
if(n<k+1 && m<k+1)
{
if(n+m-2<k)
{
printf("%d\n",-1);
return 0;
}
printf("%I64d\n",max((n/(k-m+2)),m/(k-n+2)));
return 0;
}
if(n/(k+1)*m>m/(k+1)*n)
printf("%I64d\n",n/(k+1)*m);
else
printf("%I64d\n",m/(k+1)*n);
return 0;
}
2 seconds
256 megabytes
standard input
standard output
Jzzhu is the president of country A. There are n cities numbered from 1 to n in his country. City 1 is the capital of A. Also there are mroads connecting the cities. One can go from city ui to vi (and vise versa) using the i-th road, the length of this road is xi. Finally, there are k train routes in the country. One can use the i-th train route to go from capital of the country to city si (and vise versa), the length of this route is yi.
Jzzhu doesn't want to waste the money of the country, so he is going to close some of the train routes. Please tell Jzzhu the maximum number of the train routes which can be closed under the following condition: the length of the shortest path from every city to the capital mustn't change.
The first line contains three integers n, m, k (2 ≤ n ≤ 105; 1 ≤ m ≤ 3·105; 1 ≤ k ≤ 105).
Each of the next m lines contains three integers ui, vi, xi (1 ≤ ui, vi ≤ n; ui ≠ vi; 1 ≤ xi ≤ 109).
Each of the next k lines contains two integers si and yi (2 ≤ si ≤ n; 1 ≤ yi ≤ 109).
It is guaranteed that there is at least one way from every city to the capital. Note, that there can be multiple roads between two cities. Also, there can be multiple routes going to the same city from the capital.
Output a single integer representing the maximum number of the train routes which can be closed.
5 5 3
1 2 1
2 3 2
1 3 3
3 4 4
1 5 5
3 5
4 5
5 5
2
2 2 3
1 2 2
2 1 3
2 1
2 2
2 3
2
思路：跑一遍最短路spfa，然后记录经过的边（如果普通边和铁路在某种情况下一样优则选择普通边），若其中有铁路就保留，反之不在最短路径中的铁路删除，cout删除的个数就OK了～
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
using namespace std; int n,m,k,x,y,z,cnt;
struct sdt
{
int to,len;
bool flag;
};
vector<sdt>v[100005];
long long dis[100005];
bool par[100005]; void spfa()
{
priority_queue<pair<long long,int>,vector<pair<long long,int> >,greater<pair<long long,int> > >q;
bool vis[100005]={0};
q.push(make_pair(0,1));
vis[1]=1;
while(!q.empty())
{
int s=q.top().second;
vis[s]=0;
q.pop();
for(int i=0;i<v[s].size();i++)
{
sdt p=v[s][i];
if(dis[s]!=1e18 && dis[p.to]>dis[s]+p.len)
{
dis[p.to]=dis[s]+p.len;
if(par[p.to]==1)
{
cnt--;
par[p.to]=0;
}
if(p.flag==1)
{
++cnt;
par[p.to]=1;
}
if(!vis[p.to])
{
q.push(make_pair(dis[p.to],p.to));
vis[p.to]=1;
}
}
else if(dis[s]!=1e18 && dis[p.to]==dis[s]+p.len)
{
if(!par[p.to])continue;
if(p.flag)continue;
par[p.to]=0;
cnt--;
}
}
}
} int main()
{
scanf("%d%d%d",&n,&m,&k);
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&x,&y,&z);
sdt p;
p.to=y;
p.len=z;
p.flag=0;
v[x].push_back(p);
p.to=x;
p.len=z;
p.flag=0;
v[y].push_back(p);
}
for(int i=1;i<=k;i++)
{
scanf("%d%d",&y,&z);
sdt p;
p.to=y;
p.len=z;
p.flag=1;
v[1].push_back(p);
p.to=1;
p.len=z;
p.flag=1;
v[y].push_back(p);
}
for(int i=2;i<=n;i++)
{
dis[i]=1e18;
}
spfa();
printf("%d\n",k-cnt);
return 0;
}
1 second
256 megabytes
standard input
standard output
Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to an apple store.
Jzzhu will pack his apples into groups and then sell them. Each group must contain two apples, and the greatest common divisor of numbers of the apples in each group must be greater than 1. Of course, each apple can be part of at most one group.
Jzzhu wonders how to get the maximum possible number of groups. Can you help him?
A single integer n (1 ≤ n ≤ 105), the number of the apples.
The first line must contain a single integer m, representing the maximum number of groups he can get. Each of the next m lines must contain two integers — the numbers of apples in the current group.
If there are several optimal answers you can print any of them.
6
2
6 3
2 4
9
3
9 3
2 4
6 8
2
0
思路：这是数论问题。显然若是偶数则随便组合，奇数的话，若是某质数的倍数随便组合（若在N以内此质数倍数个数为奇数，则2倍项留给偶数之后处理，否则立即匹配），当然要打标记是否用过。上述方式即可保证最优！易证。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
#include<cstring>
using namespace std; int n,cnt;
bool vis[100005];
vector<pair<int,int> >v;
bool p[100005]; void prime()
{
memset(p,1,sizeof(p));
p[1]=0;
for(int i=2;i<=n;i++)
{
if(p[i])
{
for(int j=i*2;j<=n;j+=i)
{
p[j]=0;
}
}
}
} int main()
{
scanf("%d",&n);
prime();
for(int i=3;i<=n/2;i++)
{
if(!p[i] || vis[i])continue;
int sum=0;
for(int j=1;j<=n/i;j++)if(!vis[i*j])sum++;
if(sum%2==0)
{
for(int j=1;j<=n/i;j++)
{
if(j+1>n/i)break;
if(vis[i*j])
{
continue;
}
else if(vis[i*(j+1)])
{
int k;
for(k=j+2;k<=n/i;k++)
{
if(!vis[i*k])
{
v.push_back(make_pair(j*i,i*k));
vis[j*i]=vis[i*k]=1;
++cnt;
break;
}
}
j=k;
continue;
}
v.push_back(make_pair(j*i,i*(j+1)));
vis[j*i]=vis[i*(j+1)]=1;
j++;
++cnt;
}
}
else
{
for(int j=1;j<=n/i;j++)
{
if(j+1>n/i)break;
if(j==2)continue;
if(vis[i*j])
{
continue;
}
else if(vis[i*(j+1)] || j+1==2)
{
int k;
for(k=j+2;k<=n/i;k++)
{
if(!vis[i*k])
{
v.push_back(make_pair(j*i,i*k));
vis[j*i]=vis[i*k]=1;
++cnt;
break;
}
}
j=k;
continue;
}
v.push_back(make_pair(j*i,i*(j+1)));
vis[j*i]=vis[i*(j+1)]=1;
j++;
++cnt;
}
}
} for(int i=1;i<=n/2;i++)
{
if(i+1>n/2)break;
if(vis[i*2])continue;
else if(vis[2*(i+1)])
{
int j;
for(j=i+2;j<=n/2;j++)
{
if(!vis[2*j])
{
v.push_back(make_pair(2*i,2*j));
vis[2*i]=vis[2*j]=1;
++cnt;
break;
}
}
i=j;
continue;
}
v.push_back(make_pair(2*i,2*i+2));
vis[2*i]=vis[2*(i+1)]=1;
i++;
++cnt;
} printf("%d\n",cnt);
for(int i=0;i<cnt;i++)
{
printf("%d %d\n",v[i].first,v[i].second);
}
return 0;
}