Preface
菜的批爆,B2 一直 WA 道心破碎了直接白兰去了,鉴定为纯纯的飞舞
本来想着周末补题的然后又摆了一天,E1 和 E2 都没时间补了,鉴定为纯纯的懒狗
A. Diagonals
签到,贪心枚举即可
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
int t,n,k;
signed main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%d",&t);t;--t)
{
RI i; scanf("%d%d",&n,&k); int cnt=0;
if (k==0) { puts("0"); continue; }
if (k<=n) { puts("1"); continue; }
for (cnt=1,k-=n,i=n-1;i>=1&&k>0;--i)
{
if (k>0) k-=i,++cnt;
if (k>0) k-=i,++cnt;
}
printf("%d\n",cnt);
}
return 0;
}
B1. Bouquet (Easy Version)
当时 B2 先写了发三分发现 WA 了,就赶紧滚回来把 B1 写了,这题就直接双指针扫一遍就能过
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define int long long
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=200005;
int t,n,m,a[N];
signed main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%lld",&t);t;--t)
{
RI i,j; for (scanf("%lld%lld",&n,&m),i=1;i<=n;++i)
scanf("%lld",&a[i]); sort(a+1,a+n+1);
int ans=0,sum=a[1]; for (i=j=1;i<=n;++i)
{
while (j<=n&&a[j]-a[i]<=1&&sum<=m) sum+=a[++j];
sum-=a[j--]; if (sum<=m) ans=max(ans,sum); sum-=a[i];
}
printf("%lld\n",ans);
}
return 0;
}
B2. Bouquet (Hard Version)
Div.2 B 题战俘闪总出列
这题有个很简单的讨论方式,即对于一组 \((a,a+1)\),我们假设先全部买 \(a\),然后把剩下的钱拿来买 \(a+1\)
这样如果还有剩余的钱,可以考虑将一些刚开始买的 \(a\) 替换成 \(a+1\),这种方法就不需要繁琐的讨论了
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define int long long
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=200005;
int t,n,m,a[N],b[N];
signed main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%lld",&t);t;--t)
{
RI i; for (scanf("%lld%lld",&n,&m),i=1;i<=n;++i)
scanf("%lld",&a[i]); map <int,int> rst;
for (i=1;i<=n;++i) scanf("%lld",&b[i]),rst[a[i]]=b[i];
int ans=0;
for (auto [a,x]:rst)
{
int c1=min(x,m/a),sum=c1*a,rem=m-c1*a;
if (!rst.count(a+1)) { ans=max(ans,sum); continue; }
int c2=min(rst[a+1],rem/(a+1));
sum+=c2*(a+1); rem-=c2*(a+1);
sum+=min(c1,min(rem,rst[a+1]-c2));
ans=max(ans,sum);
}
printf("%lld\n",ans);
}
return 0;
}
C. Squaring
神秘题,感觉精度要爆炸但交上去就是过了,很神秘
首先这题的策略很简单,即从前往后贪心,每个位置操作到比前一个位置大了就停手一定最优
现在的难点就在于怎么比较两个数的大小,假设前一个数 \(a_{i-1}\) 操作了 \(x\) 次,考虑快速计算当前数 \(a_{i}\) 操作的次数 \(y\)
原来要比较 \(a_{i-1}^{2^x}\) 和 \(a_i^{2^y}\) 的大小;两边取 \(\ln\) 后变为比较 \(2^x\times \ln a_{i-1}\) 和 \(2^y\times \ln a_{i}\) 的大小
两边再做商得 \(2^{x-y}\times \frac{\ln a_{i-1}}{\ln a_i}\),因此可以直接再取 \(\log_2\) 快速解出 \(y\) 的值,注意要对 \(0\) 取 \(\max\)
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define int long long
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=200005;
int t,n,m,a[N];
signed main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%lld",&t);t;--t)
{
RI i,j; for (scanf("%lld",&n),i=1;i<=n;++i) scanf("%lld",&a[i]);
int ans=0,lst=0; bool flag=1;
for (i=2;i<=n;++i)
{
if (a[i]==1&&a[i-1]!=1) { flag=0; break; }
lst=max(0LL,(int)ceil(log2(1.0L*log(a[i-1])/log(a[i]))+lst));
ans+=lst;
}
if (!flag) { puts("-1"); continue; }
printf("%lld\n",ans);
}
return 0;
}
D. Cases
考虑原题的限制相当于对于任意连续的 \(k\) 个字符,其中必须至少有一个是终止字符
反过来思考下,这就等价于对于连续的 \(k\) 个字符组成的状压状态 \(T\) 不能为所有不是终止字符的字符构造的集合 \(S\) 的子集
因此把所有连续的 \(k\) 个字符对应的状态标记为不合法(注意最后一个字符要特判),然后把它们的超集也标记成不合法即可
最后用枚举子集或者 sosdp
即可 \(O(3^{c})\) 或 \(O(2^{c}\times c)\) 通过此题
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=(1<<18)+5;
int t,n,c,k,f[N],bkt[20]; char s[N];
signed main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%d",&t);t;--t)
{
RI i,j; scanf("%d%d%d%s",&n,&c,&k,s+1);
for (i=0;i<(1<<c);++i) f[i]=0; f[1<<(s[n]-'A')]=1;
int mask=0; for (i=0;i<c;++i) bkt[i]=0;
auto add=[&](CI x)
{
if (++bkt[x]==1) mask^=(1<<x);
};
auto del=[&](CI x)
{
if (--bkt[x]==0) mask^=(1<<x);
};
for (i=1;i<=k;++i) add(s[i]-'A'); f[mask]=1;
for (i=k+1;i<=n;++i) del(s[i-k]-'A'),add(s[i]-'A'),f[mask]=1;
for (i=0;i<c;++i) for (j=0;j<(1<<c);++j)
if ((j>>i)&1) f[j]|=f[j^(1<<i)];
int ans=c; for (i=0;i<(1<<c);++i)
if (!f[i]) ans=min(ans,__builtin_popcount(((1<<c)-1)^i));
printf("%d\n",ans);
}
return 0;
}
Postscript
感觉再这么菜下去要被新一届 Div2 薄纱了
标签:typedef,961,int,Codeforces,long,pair,Div,include,define From: https://www.cnblogs.com/cjjsb/p/18328832