Codeforces Round #570 Div. 3
A：暴力从小到大枚举判断。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
int n=read();
for (int i=n;i<=n+10000;i++)
{
int s=0,x=i;
while (x) s+=x%10,x/=10;
if (s%4==0) {cout<<i;return 0;}
}
return 0;
//NOTICE LONG LONG!!!!!
}
B：显然max-min<=2k时有解，最优解为min+k。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 110
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int q,n,k,a[N];
signed main()
{
q=read();
while (q--)
{
n=read(),k=read();
for (int i=1;i<=n;i++) a[i]=read();
int mn=inf,mx=0;
for (int i=1;i<=n;i++) mn=min(mn,a[i]),mx=max(mx,a[i]);
if (mx-k<=mn+k) cout<<mn+k<<endl;
else cout<<-1<<endl;
}
return 0;
//NOTICE LONG LONG!!!!!
}
C：一点脑子都不想动可以二分答案。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 110
#define int long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int q,k,n,a,b;
signed main()
{
q=read();
while (q--)
{
k=read()-1,n=read(),a=read(),b=read();
if (b*n>k) printf("-1\n");
else
{
int l=0,r=n,ans=0;
while (l<=r)
{
int mid=l+r>>1;
if (a*mid+b*(n-mid)<=k) ans=mid,l=mid+1;
else r=mid-1;
}
printf("%d\n",ans);
}
}
return 0;
//NOTICE LONG LONG!!!!!
}
D、E：跳过。
F：当只要求选择两个数时，显然选择最大值及最大的不是其因子的数最优。因为假设次大值是最大值的因子，那么次大值<=最大值/2，如果不选择最大值，两个数的和一定不大于最大值。于是考虑枚举最大值，剩下的数按上述方法选择。可以使用set删去因子，复杂度O(nsqrt(n)log(n))但也能跑过。实际上sort后枚举最大值并暴力找去掉因子的最大数复杂度就是O(nsqrt(n))。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int q,n,a[N],b[N],t;
set<int> f;
void del(int x)
{
f.erase(x);
b[++t]=x;
}
signed main()
{
q=read();
while (q--)
{
n=read();
for (int i=1;i<=n;i++) a[i]=read();f.clear();
sort(a+1,a+n+1);n=unique(a+1,a+n+1)-a-1;
int ans=0;
for (int i=1;i<=n;i++) f.insert(a[i]);
for (int i=n;i>=1;i--)
{
t=0;
int s=a[i];f.erase(a[i]);
for (int j=1;j*j<=a[i];j++)
if (a[i]%j==0)
{
if (f.find(j)!=f.end()) del(j);
if (f.find(a[i]/j)!=f.end()) del(a[i]/j);
}
if (!f.empty())
{
int x=*(--f.end());
for (int j=1;j*j<=x;j++)
if (x%j==0)
{
if (f.find(j)!=f.end()) del(j);
if (f.find(x/j)!=f.end()) del(x/j);
}
s+=x;
}
if (!f.empty()) s+=*(--f.end());
ans=max(ans,s);
for (int j=1;j<=t;j++) f.insert(b[j]);
}
printf("%d\n",ans);
}
return 0;
//NOTICE LONG LONG!!!!!
}
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int q,n,a[N];
signed main()
{
q=read();
while (q--)
{
n=read();
for (int i=1;i<=n;i++) a[i]=read();
sort(a+1,a+n+1);n=unique(a+1,a+n+1)-a-1;
int ans=0;
for (int i=n;i>=1;i--)
{
int s=a[i];int x=0;
for (int j=i-1;j>=1;j--)
if (a[i]%a[j]) {x=j,s+=a[j];break;}
for (int j=x-1;j>=1;j--)
if (a[i]%a[j]&&a[x]%a[j]) {s+=a[j];break;}
ans=max(ans,s);
}
printf("%d\n",ans);
}
return 0;
//NOTICE LONG LONG!!!!!
}
G：求出每种糖果数量以及其中有多少个fi=1。然后从大到小考虑礼物中的糖果数量，将糖果数量足够的糖果种类加入set，取出Σfi最大的。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int q,n;
struct data
{
int x,y;
bool operator <(const data&a) const
{
return x<a.x||x==a.x&&y<a.y;
}
}a[N],c[N];
multiset<int> f;
signed main()
{
q=read();
while (q--)
{
n=read();
for (int i=1;i<=n;i++) a[i].x=read(),a[i].y=read();
sort(a+1,a+n+1);
int m=0;
for (int i=1;i<=n;i++)
{
int t=i;
while (t<n&&a[t+1].x==a[i].x) t++;
c[++m].x=t-i+1;c[m].y=0;
for (int j=i;j<=t;j++) if (a[j].y==1) c[m].y++;
i=t;
}
sort(c+1,c+m+1);reverse(c+1,c+m+1);
int cur=0,ans=0,ans2=0;f.clear();
for (int i=n;i>=1;i--)
{
while (cur<m&&c[cur+1].x>=i)
{
f.insert(c[++cur].y);
}
if (!f.empty())
{
ans+=i;
ans2+=min(i,*(--f.end()));
f.erase(--f.end());
}
}
printf("%d %d\n",ans,ans2);
}
return 0;
//NOTICE LONG LONG!!!!!
}
H：建出序列自动机（对每个位置求出每个字符下一次出现位置），然后dp出长度为i的本质不同子序列数量即可。注意数量对k取min。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 110
#define int long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,f[N][N],nxt[N][26];
ll k,ans;
char s[N];
signed main()
{
n=read();ll k;cin>>k;
scanf("%s",s+1);
ans=k*n;
for (int i=0;i<26;i++) nxt[n+1][i]=n+1;
for (int i=n;i>=1;i--)
{
for (int j=0;j<26;j++) nxt[i][j]=nxt[i+1][j];
nxt[i][s[i]-'a']=i;
}
for (int i=0;i<=n;i++) f[i][1]=1;
for (int i=2;i<=n+1;i++)
for (int j=n;j>=0;j--)
{
for (int k=0;k<26;k++)
f[j][i]+=f[nxt[j+1][k]][i-1];
f[j][i]=min(f[j][i],k);
}
for (int i=n+1;i>=1;i--)
{
if (k>f[0][i]) ans-=(i-1)*f[0][i],k-=f[0][i];
else {ans-=(i-1)*k;k=0;break;}
}
if (k) cout<<-1;else
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
小号5。result：rank 2 rating +322
