2016 ACM/ICPC Asia Regional Qingdao Online（部分题解）
5878---I Count Two Three http://acm.hdu.edu.cn/showproblem.php?pid=5878
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1287 Accepted Submission(s): 604
It all started several months ago.
We found out the home address of the enlightened agent Icount2three and decided to draw him out.
Millions of missiles were detonated, but some of them failed.
After the event, we analysed the laws of failed attacks.
It's interesting that the i-th attacks failed if and only if i can be rewritten as the form of 2a3b5c7d which a,b,c,d are non-negative integers.
At recent dinner parties, we call the integers with the form 2a3b5c7d "I Count Two Three Numbers".
A related board game with a given positive integer n from one agent, asks all participants the smallest "I Count Two Three Number" no smaller than n.
1
11
13
123
1234
12345
123456
1234567
12345678
123456789
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std; int t,n,cnt;
long long st[500005],p,r;
long long p2[500005],p3[500005],p5[500005],p7[500005]; int main()
{
scanf("%d",&t);
r=1;
p2[0]=r;
for(int i=1;i;i++)
{
if(r>1e9)break;
r*=2;
p2[i]=r;
}
r=1;
p3[0]=r;
for(int i=1;i;i++)
{
if(r>1e9)break;
r*=3;
p3[i]=r;
}
r=1;
p5[0]=r;
for(int i=1;i;i++)
{
if(r>1e9)break;
r*=5;
p5[i]=r;
}
r=1;
p7[0]=r;
for(int i=1;i;i++)
{
if(r>1e9)break;
r*=7;
p7[i]=r;
}
cnt=0;
for(int i=0;i<=31;i++)
{
for(int j=0;j<=19;j++)
{
for(int k=0;k<=12;k++)
{
for(int v=0;v<=11;v++)
{
p=p2[i]*p3[j];
if(p>1e9)break;
st[++cnt]=p;
if(p<1e9)
{
p*=p5[k];
if(p>1e9)break;
st[++cnt]=p;
}
else break;
if(p<1e9)
{
p*=p7[v];
if(p>1e9)break;
st[++cnt]=p;
}
}
}
}
}
sort(st+1,st+cnt+1);
while(t--)
{
scanf("%d",&n);
printf("%lld\n",*lower_bound(st+1,st+cnt+1,n));
}
return 0;
}
5879---Cure http://acm.hdu.edu.cn/showproblem.php?pid=5879
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1783 Accepted Submission(s): 568
For each test case, there is a single line, containing a single positive integer n.
The input file is at most 1M.
2
4
8
15
1.25000
1.42361
1.52742
1.58044
这题可以找规律发现到某一点往上至正无穷大，数值都不会变，然后就好做啦～只需要处理前几十万个。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
using namespace std; string n;
long double sum[400005]; int main()
{
sum[0]=(long double)0.0;
for(int i=1;i<=400000;i++)
{
sum[i]=sum[i-1]+(long double)1.0/(long double)i/(long double)i;
}
while(cin>>n)
{
if(n[0]=='-')n[0]='0';
long long p=0;
bool f=0;
for(int i=0;i<n.size();i++)
{
p=p*10+n[i]-48;
if(p>400000)
{
printf("%.5f\n",(double)sum[400000]);
f=1;
break;
}
}
if(f==0)
{
printf("%.5f\n",(double)sum[p]);
}
}
return 0;
}
5882---Balanced Game http://acm.hdu.edu.cn/showproblem.php?pid=5882
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 435 Accepted Submission(s): 377
Recently, there is a upgraded edition of this game: rock-paper-scissors-Spock-lizard, in which there are totally five shapes. The rule is simple: scissors cuts paper; paper covers rock; rock crushes lizard; lizard poisons Spock; Spock smashes scissors; scissors decapitates lizard; lizard eats paper; paper disproves Spock; Spock vaporizes rock; and as it always has, rock crushes scissors.
Both rock-paper-scissors and rock-paper-scissors-Spock-lizard are balanced games. Because there does not exist a strategy which is better than another. In other words, if one chooses shapes randomly, the possibility he or she wins is exactly 50% no matter how the other one plays (if there is a tie, repeat this game until someone wins). Given an integer N, representing the count of shapes in a game. You need to find out if there exist a rule to make this game balanced.
For each test case, there is only one line with an integer N (2≤N≤1000), as described above.
Here is the sample explanation.
In the first case, donate two shapes as A and B. There are only two kind of rules: A defeats B, or B defeats A. Obviously, in both situation, one shapes is better than another. Consequently, this game is not balanced.
In the second case, donate two shapes as A, B and C. If A defeats B, B defeats C, and C defeats A, this game is balanced. This is also the same as rock-paper-scissors.
In the third case, it is easy to set a rule according to that of rock-paper-scissors-Spock-lizard.
2
3
5
Balanced
Balanced
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
using namespace std; int t,n;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
if(n%2==0)cout<<"Bad"<<endl; else cout<<"Balanced"<<endl;
}
return 0;
}
5883---The Best Pathhttp://acm.hdu.edu.cn/showproblem.php?pid=5883
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1155 Accepted Submission(s): 515
For each test case, in the first line there are two positive integers N (N≤100000) and M (M≤500000), as described above. The i-th line of the next N lines contains an integer ai(∀i,0≤ai≤10000) representing the number of the i-th lake.
The i-th line of the next M lines contains two integers ui and vi representing the i-th river between the ui-th lake and vi-th lake. It is possible that ui=vi.
3 2
3
4
5
1 2
2 3
4 3
1
2
3
4
1 2
2 3
2 4
裸的一笔画问题，判欧拉回路与欧拉通路两种与连通性。
欧拉通路没啥好说的，首尾点固定，路线固定。
欧拉回路，即环。值是不定的，要求最大，即将必须通过的点XOR，再枚举开头初始点，与之前ans进行XOR，比最值。
不是一笔画即Impossible。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
using namespace std; int t,n,m,x,y,ans,p,cnt;
int a[100005],sum[100005],par[100005]; int find(int x)
{
if(x==par[x])return x; else return par[x]=find(par[x]);
} void unionx(int x,int y)
{
x=find(x);
y=find(y);
if(x==y)return ;
par[x]=y;
} int main()
{
scanf("%d",&t);
while(t--)
{
for(int i=1;i<=n;i++)par[i]=i;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
memset(sum,0,sizeof(sum));
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
unionx(x,y);
sum[x]++;
sum[y]++;
}
cnt=0;
for(int i=1;i<=n;i++)
{
if(i==par[i] && sum[i]!=0)cnt++;
}
if(cnt>1)
{
cout<<"Impossible"<<endl;
continue;
}
p=0;
ans=0;
for(int i=1;i<=n;i++)
{
if(sum[i]==0)continue;
if(sum[i]%2==1)p++;
ans^=(((sum[i]+1)/2)%2)*a[i];
}
if(p!=0 && p!=2)
{
cout<<"Impossible"<<endl;
continue;
}
if(p==2)
{
printf("%d\n",ans);
continue;
}
else
{
int anss=0;
for(int i=1;i<=n;i++)
{
if(sum[i]==0)continue;
anss=max(anss,ans^a[i]);
}
printf("%d\n",anss);
}
}
return 0;
}
先只看了几道通过率高的题A了，剩余的题如果还有A的，且心情好的话再补上。。。
毕竟之前两天辣么的题目都没挂上来。原因。。。文化课掉出年级250外了。。。