牛客小白月赛69 ABCDE
https://ac.nowcoder.com/acm/contest/52441
这场小白我给打的，我愿称之为年度喜剧片
A-蛋挞
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN;
const LL N=2e6+10,M=4004;
#define endl '\n'
LL a[N];
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
//cin>>T;
while(T--)
{
LL a,b;
cin>>a>>b;
LL sum1=a/b;
LL sum2=a%b;
if(sum2>sum1) cout<<"niuniu eats more than others"<<endl;
else if(sum2<sum1) cout<<"niuniu eats less than others"<<endl;
else cout<<"same"<<endl;
}
return 0;
}
B-玩具
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN;
const LL N=2e6+10,M=4004;
#define endl '\n'
LL a[N];
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
//cin>>T;
while(T--)
{
LL n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
sort(a+1,a+1+n);
LL sum=0;
if(n%2==0)
{
for(int i=2;i<=n;i+=2)
{
sum+=a[i];
}
}
else
{
for(int i=1;i<=n;i+=2)
{
sum+=a[i];
}
}
cout<<sum<<endl;
}
return 0;
}
C-开题顺序
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN;
const LL N=2e6+10,M=4004;
#define endl '\n'
LL n,t,p;
LL maxn=0,idx[N];
LL a[N],b[N],c[N],x[N],y[N];
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
//cin>>T;
while(T--)
{
cin>>n>>t>>p;
for(int i=1;i<=n;i++)
{
idx[i]=i;
}
for(int i=1;i<=n;i++)
{
cin>>a[i]>>b[i]>>c[i]>>x[i]>>y[i];
}
do
{
LL sum=0;
LL time=0;
for(int i=1;i<=n;i++)
{
time+=x[idx[i]];
if(time<=t)
{
sum+=max(c[idx[i]],a[idx[i]]-time*b[idx[i]]-y[idx[i]]*p);
}
else break;
}
maxn=max(maxn,sum);
}
while(next_permutation(idx+1,idx+1+n));
cout<<maxn<<endl;
}
return 0;
}
D-旅游
输入
4 6 7
1 2 3
1 3 4
1 4 6
2 3 2
2 4 1
3 4 5
输出
1
说实话这题我好像很久很久之前做过一个类似的，思路也差不多
就是先把修复的价钱按照从小到大排个序
前面几个如果n个点都已经在了的话，就选取前面几个价钱就行
然后这几个价钱从大到小加起来，哪里超过了牛牛的承受范围就停止丢给国家去修
（谁能想到一条语句放错了位置我写了一个半小时，人生疑惑）
妥妥 年度喜剧片
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN;
const LL N=2e6+10,M=2023;
const LL mod=998244353;
const double PI=3.1415926535;
#define endl '\n'
LL father[N],siz[N];
LL n,m,k,cnt=0;
struct node
{
LL x,y,z;
}flag[N];
LL sum[N];
bool cmp(node l,node r)
{
return l.z<r.z;
}
void init()
{
for(int i=1;i<=n;i++)
{
father[i]=i;
siz[i]=1;
}
}
LL find(LL x)
{
if(father[x]!=x) father[x]=find(father[x]);
return father[x];
}
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
//cin>>T;
while(T--)
{
cin>>n>>m>>k;
init();
for(int i=1;i<=m;i++)
{
cin>>flag[i].x>>flag[i].y>>flag[i].z;
}
sort(flag+1,flag+1+m,cmp);
for(int i=1;i<=m;i++)
{
LL fx=find(flag[i].x);
LL fy=find(flag[i].y);
if(fx!=fy)
{
father[max(fx,fy)]=min(fx,fy);
siz[min(fx,fy)]+=siz[max(fx,fy)];
sum[++cnt]=flag[i].z;
}
//cout<<siz[min(fx,fy)]<<endl;
if(siz[min(fx,fy)]==n) break;
}
LL idx=-1;
LL res=0;
sum[0]=0;
for(int i=cnt;i>=1;i--)
{
//cout<<sum[i]<<" ";
res+=(cnt-i+1)*sum[i];
if(res>k)
{
idx=i;
break;
}
}
if(idx==-1) cout<<"0"<<endl;
else cout<<sum[idx]<<endl;
}
return 0;
}
E-等腰三角形(easy)
真的会有人连两边之和大于第三边都忘记吗？
真的吗真的吗真的有人吗？
是我了┭┮﹏┭┮（傻狗爆哭）
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN;
const LL N=2e6+10,M=2023;
const LL mod=998244353;
const double PI=3.1415926535;
#define endl '\n'
LL n;
double a[N],b[N];
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
int T=1;
//cin>>T;
while(T--)
{
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i]>>b[i];
}
LL sum=0;
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
for(int k=j+1;k<=n;k++)
{
if((a[i]==a[j]&&a[j]==a[k])||(b[i]==b[j]&&b[j]==b[k])) continue;
//AB∣=√ [ (x1－x2)²+ (y1－y2)²]
double num1=sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
double num2=sqrt((a[i]-a[k])*(a[i]-a[k])+(b[i]-b[k])*(b[i]-b[k]));
double num3=sqrt((a[k]-a[j])*(a[k]-a[j])+(b[k]-b[j])*(b[k]-b[j]));
//cout<<num1<<" "<<num2<<" "<<num3<<endl;
if(num1+num2>num3&&num1+num3>num2&&num2+num3>num1)
{
if(num1==num2||num2==num3||num1==num3) sum++;
}
}
}
}
cout<<sum<<endl;
}
return 0;
}
