Topic link:
http://codeforces.com/contest/1154
A. Restoring Three Numbers
Attention should be paid to the meaning of the title: a+b is not - a+b;
The code is as follows:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e2+5;
int n;
ll x[4];
int main()
{
for (int i=0;i<4;i++)
{
scanf("%lld",&x[i]);
}
int loc=-1;
for (int i=0;i<4;i++)
{
if(2*x[i]==(x[(i+1)%4]+x[(i+2)%4]+x[(i+3)%4]))
{
loc=i;
break;
}
}
printf("%lld %lld %lld\n",x[loc]-x[(loc+1)%4],x[loc]-x[(loc+2)%4],x[loc]-x[(loc+3)%4]);
return 0;
}B. Make Them Equal
Calculate the types of different numbers, and then discuss them by classification.
If type > 3, direct output - 1
If type== 3, see if the absolute difference between left and right numbers is equal to the absolute difference between the middle numbers.
If kind==2, then there must be a solution, because the minimum, then try to narrow the scope, if the difference is even, then the answer needs / 2, if it is odd, then output directly.
If type== 0, output 0
The code is as follows:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e2+5;
int n;
int a[maxn];
int num[maxn];
int main()
{
int kind=0;
memset (num,0,sizeof(num));
scanf("%d",&n);
for (int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(num[a[i]]==0)
{
kind++;
}
num[a[i]]++;
}
if(kind==1)
{
printf("0\n");
return 0;
}
if(kind==2)
{
int cnt=0;
int b[5];
for (int i=0;i<=100;i++)
{
if(num[i])
{
b[cnt++]=i;
}
}
if((b[0]+b[1])%2)
{
printf("%d\n",abs(b[0]-b[1]));
}
else
printf("%d\n",abs(b[0]-b[1])/2);
return 0;
}
if(kind==3)
{
int cnt=0;
int b[5];
for (int i=0;i<=100;i++)
{
if(num[i])
{
b[cnt++]=i;
}
}
if(b[2]-b[1]==b[1]-b[0])
{
printf("%d\n",b[1]-b[0]);
}
else printf("-1\n");
}
if(kind>=4) printf("-1\n");
return 0;
}C. Gourmet Cat
The simulation starts at one day of the week, then starts to simulate, and then finds the maximum value.
The code is as follows:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e2+5;
int n;
ll numa,numb,numc;
ll ans=-1;
int main()
{
scanf("%lld%lld%lld",&numa,&numb,&numc);
for (int i=1;i<=7;i++)
{
ll tnuma=numa,tnumb=numb,tnumc=numc;
ll temp=0;
int ok=0;
for (int j=i;j<=7;j++)
{
if((j==1||j==4||j==7))
{
if(tnuma)
{
tnuma--;
temp++;
}
else
{
ok=1;
break;
}
}
else if(j==2||j==6)
{
if(tnumb)
{
tnumb--;
temp++;
}
else
{
ok=1;
break;
}
}
else if(j==3||j==5)
{
if(tnumc)
{
tnumc--;
temp++;
}
else
{
ok=1;
break;
}
}
}
if(ok==1)
{
ans=max(ans,temp);
continue;
}
ll t1=tnuma/3;
ll t2=tnumb/2;
ll t3=tnumc/2;
ll ci=min(t1,min(t2,t3));
tnuma=tnuma-ci*3;
tnumb=tnumb-ci*2;
tnumc=tnumc-ci*2;
temp=temp+ci*7;
for (int j=1;j<=7;j++)
{
if((j==1||j==4||j==7))
{
if(tnuma)
{
tnuma--;
temp++;
}
else
{
break;
}
}
else if(j==2||j==6)
{
if(tnumb)
{
tnumb--;
temp++;
}
else
{
break;
}
}
else if(j==3||j==5)
{
if(tnumc)
{
tnumc--;
temp++;
}
else
{
break;
}
}
}
ans=max(ans,temp);
// printf("i=%d ans=%lld\n",i,ans);
}
printf("%lld\n",ans);
return 0;
}D. Walking Robot
Greedy, note that the maximum number of batteries in this question is storage capacity, can not exceed.
When s[i]=0, the batteries are preferred.
When s[i]=1, if the battery i s full at this time, the battery should be used first, and then the battery should be used. Note that the battery + 1 can not exceed its capacity.
The code is as follows:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=2*1e6+5;
int n,a,b;
int s[maxn];
int main()
{
scanf("%d%d%d",&n,&b,&a);
int nb=b,na=a;
for (int i=1;i<=n;i++)
{
scanf("%d",&s[i]);
}
int ans=0;
for (int i=1;i<=n;i++)
{
if(s[i]==0)
{
if(a==na)
{
ans++;
na--;
}
else if(na>0)
{
ans++;
na--;
}
else
{
if(nb>0)
{
ans++;
nb--;
}
else
{
break;
}
}
}
else
{
if(a==na)
{
ans++;
na--;
}
else if(nb>0)
{
nb--;
na++;
ans++;
}
else
{
if(na>0)
{
na--;
ans++;
}
else
{
break;
}
}
}
}
printf("%d\n",ans);
return 0;
}E. Two Teams
Set up a two-way linked list to solve this problem.
The code is as follows:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=2*1e5+5;
int n,k;
int loc[maxn];
int vis[maxn];
int ans[maxn];
struct List
{
int next;
int pre;
int data;
};
List l [maxn];
int main()
{
scanf("%d%d",&n,&k);
memset (vis,0,sizeof(vis));
memset (l,-1,sizeof(l));
for (int i=1;i<=n;i++)
{
int x;
scanf("%d",&x);
loc[x]=i;
l[i].data=x;
if(i==1)
{
l[i].pre=-1;
continue;
}
l[i].pre=i-1;
l[i-1].next=i;
}
l[n].next=-1;
int num=0;
int kt=n;
int ci=0;
while(num<n)
{
ci++;
int an;
if(ci%2) an=1;
else an=2;
while(vis[kt]) kt--;
int pos=loc[kt];
vis[kt]=1;
ans[pos]=an;
num++;
int hpo=pos;
int tpo=pos;
int hp=-1,tp=-1;
for (int i=0;i<k;i++)
{
if(l[hpo].pre==-1)
{
hp=-1;
break;
}
hpo=l[hpo].pre;
ans[hpo]=an;
vis[l[hpo].data]=1;
hp=hpo;
num++;
}
for (int i=0;i<k;i++)
{
if(l[tpo].next==-1)
{
tp=-1;
break;
}
tpo=l[tpo].next;
ans[tpo]=an;
vis[l[tpo].data]=1;
tp=tpo;
num++;
}
if(hp==-1&&tp==-1) break;
if(hp!=-1&&tp!=-1)
{
if(l[tp].next!=-1) l[l[tp].next].pre=l[hp].pre;
if(l[hp].pre!=-1) l[l[hp].pre].next=l[tp].next;
}
if(hp==-1)
{
if(l[tp].next!=-1) l[l[tp].next].pre=-1;
}
if(tp==-1)
{
if(l[hp].pre!=-1) l[l[hp].pre].next=-1;
}
//printf("num=%d\n",num);
}
for (int i=1;i<=n;i++) printf("%d",ans[i]);
printf("\n");
return 0;
}