【比赛记录】ARC133
复健训练了属于是））
考虑直接找第一个不上升的位置。如果没有就直接选最后一个数。
#include
using namespace std;
typedef double db;
#define int long long
#define fi first
#define se second
#define mk make_pair
#define pb emplace_back
#define poly vector
#define Bt(a) bitset
#define bc __builtin_popcount
#define pc putchar
#define ci const int&
const int mod = 1e9 + 7;
const db eps = 1e-10;
const int inf = (1 << 30);
inline int Max(ci x, ci y) {return x > y ? x : y;}
inline int Min(ci x, ci y) {return x < y ? x : y;}
inline db Max(db x, db y) {return x - y > eps ? x : y;}
inline db Min(db x, db y) {return x - y < eps ? x : y;}
inline int Add(ci x, ci y, ci M = mod) {return (x + y) % M;}
inline int Mul(ci x, ci y, ci M = mod) {return 1ll * x * y % M;}
inline int Dec(ci x, ci y, ci M = mod) {return (x - y + M) % M;}
typedef pair pii;
inline int Abs(int x) {return x < 0 ? -x : x;}
//char buf[1<<21],*p1=buf,*p2=buf;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char Obuf[105000],*O=Obuf;//Siz shoule be the size of Out File
int pst[30],ptop;
inline void Fprint(){fwrite(Obuf,1,O-Obuf,stdout);}
inline void Fwrite(int x){
if(x==0){*O++='0';return;}
if(x<0)*O++='-',x=-x;ptop=0;
while(x)pst[++ptop]=x%10,x/=10;
while(ptop)*O++=pst[ptop--]+'0';
if(O-Obuf>100000)Fprint(),O=Obuf;
}
inline int read() {
int s = 0, w = 1;
char ch = getchar();
while (!isdigit(ch)) {if (ch == '-') w = -1;ch = getchar();}
while (isdigit(ch)) {s = s * 10 + ch - '0';ch = getchar();}
return s * w;
}
inline void write(int x) {
if (x < 0)putchar('-'), x = -x;
if (x > 9)write(x / 10);
pc(x % 10 + '0');
}
inline int qpow(int x, int y) {
int res = 1;
while (y) {if (y & 1)res = Mul(res, x);x = Mul(x, x);y >>= 1;}
return res;
}
inline void cadd(int &x, int y) {x += y;}
inline void cmul(int &x, int y) {x *= y;}
inline void cmax(int &x, int y) {x = Max(x, y);}
inline void cmin(int &x, int y) {x = Min(x, y);}
const int N = 2e5 + 10;
namespace Refined_heart{
int n,a[N];
void solve(){
n=read();
for(int i=1;i<=n;++i)a[i]=read();
int pre=-1;
for(int i=1;i<=n;++i){
if(pre>a[i]){
break;
}
cmax(pre,a[i]);
}
if(pre==-1)pre=a[n];
poly A;
for(int i=1;i<=n;++i)if(a[i]!=pre)A.pb(a[i]);
for(auto v:A)write(v),pc(' ');
}
}
signed main(){
Refined_heart::solve();
return 0;
}
仔细读题，这是排列。
考虑暴力找到每个 \(A_i\) 的倍数，然后直接设 \(f_{i,j}\) 表示 \(A_i\) 从位置 \(j\) 转移过来的方案数。
这是因为，我们的总有效状态只有 \(\sum \frac{n}{i}=n\ln n\) 个。
然后考虑转移。设我们要匹配的位置是 \(now,\) 那么之前所有值的匹配位置都不能超过 \(now.\)
所以找一个数据结构来维护前缀最大值以及单点修改即可。这里写了一个线段树。
#include
using namespace std;
typedef double db;
#define int long long
#define fi first
#define se second
#define mk make_pair
#define pb emplace_back
#define poly vector
#define Bt(a) bitset
#define bc __builtin_popcount
#define pc putchar
#define ci const int&
const int mod = 1e9 + 7;
const db eps = 1e-10;
const int inf = (1 << 30);
inline int Max(ci x, ci y) {return x > y ? x : y;}
inline int Min(ci x, ci y) {return x < y ? x : y;}
inline db Max(db x, db y) {return x - y > eps ? x : y;}
inline db Min(db x, db y) {return x - y < eps ? x : y;}
inline int Add(ci x, ci y, ci M = mod) {return (x + y) % M;}
inline int Mul(ci x, ci y, ci M = mod) {return 1ll * x * y % M;}
inline int Dec(ci x, ci y, ci M = mod) {return (x - y + M) % M;}
typedef pair pii;
inline int Abs(int x) {return x < 0 ? -x : x;}
//char buf[1<<21],*p1=buf,*p2=buf;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char Obuf[105000],*O=Obuf;//Siz shoule be the size of Out File
int pst[30],ptop;
inline void Fprint(){fwrite(Obuf,1,O-Obuf,stdout);}
inline void Fwrite(int x){
if(x==0){*O++='0';return;}
if(x<0)*O++='-',x=-x;ptop=0;
while(x)pst[++ptop]=x%10,x/=10;
while(ptop)*O++=pst[ptop--]+'0';
if(O-Obuf>100000)Fprint(),O=Obuf;
}
inline int read() {
int s = 0, w = 1;
char ch = getchar();
while (!isdigit(ch)) {if (ch == '-') w = -1;ch = getchar();}
while (isdigit(ch)) {s = s * 10 + ch - '0';ch = getchar();}
return s * w;
}
inline void write(int x) {
if (x < 0)putchar('-'), x = -x;
if (x > 9)write(x / 10);
pc(x % 10 + '0');
}
inline int qpow(int x, int y) {
int res = 1;
while (y) {if (y & 1)res = Mul(res, x);x = Mul(x, x);y >>= 1;}
return res;
}
inline void cadd(int &x, int y) {x += y;}
inline void cmul(int &x, int y) {x *= y;}
inline void cmax(int &x, int y) {x = Max(x, y);}
inline void cmin(int &x, int y) {x = Min(x, y);}
const int N = 2e6 + 10;
namespace Refined_heart{
int n,a[N],b[N];
int pos[N];
unordered_mapf[N];
namespace SGT{
int maxn[N];
void build(int l,int r,int x){
if(l==r)return;
int mid=(l+r)>>1;
build(l,mid,x<<1);
build(mid+1,r,x<<1|1);
}
inline void pushup(int x){
cmax(maxn[x],Max(maxn[x<<1],maxn[x<<1|1]));
}
void change(int x,int l,int r,int pos,int v){
if(l==r){
cmax(maxn[x],v);
return;
}
int mid=(l+r)>>1;
if(pos<=mid)change(x<<1,l,mid,pos,v);
else change(x<<1|1,mid+1,r,pos,v);
pushup(x);
}
int query(int x,int L,int R,int l,int r){
if(L>=l&&R<=r)return maxn[x];
int ans=-1,mid=(L+R)>>1;
if(l<=mid)ans=Max(ans,query(x<<1,L,mid,l,r));
if(midnxt;
void solve(){
n=read();build(1,n,1);
for(int i=1;i<=n;++i)a[i]=read();
for(int i=1;i<=n;++i)b[i]=read();
for(int i=1;i<=n;++i)pos[b[i]]=i;
for(int i=a[1];i<=n;i+=a[1]){
f[1][pos[i]]=1;
change(1,1,n,pos[i],1);
}
for(int i=2;i<=n;++i){
for(int j=a[i];j<=n;j+=a[i]){
int now=pos[j];
if(now==1)f[i][now]=1;
else f[i][now]=query(1,1,n,1,now-1)+1;
nxt.pb(mk(now,f[i][now]));
}
for(auto v:nxt){
int ps=v.fi;
int vl=v.se;
change(1,1,n,ps,vl);
}
nxt.clear();
}
int ans=-1;
for(int i=1;i<=n;++i){
for(int j=a[i];j<=n;j+=a[i]){
cmax(ans,f[i][pos[j]]);
}
}
write(ans);
}
}
signed main(){
Refined_heart::solve();
return 0;
}
贪心神题。
首先考虑将所有位置填成 \(k-1,\) 再进行调整。
那么对于 \(a_i,\) 这一行要修改的数自然就是 \(m\times (k-1)-a_i.\) 这是最少的修改次数。
设所有行的修改数的和为 \(C_i,\) 列的为 \(D_i,\) 那么我们考虑证明答案是 \(n\times m\times (k-1)-\max\{\sum c,\sum d\}\)
首先，无解当且仅当 \(\sum a\not\equiv \sum b\) 这是因为两侧都是所有数的和。
那么，考虑给 \(c,d\) 中较小的加上一个 \(k\) 的倍数使得二者相等。
那么考虑重复下列操作直到矩阵为 \(0\)：
- 找到一个位置满足 \(c_i>0,d_i>0\) 并令二者减一
在上述的调整上，没有一个数会被修改为负数。而修改次数就恰好是 \(\max\{\sum c,\sum d\}\)
#include
using namespace std;
typedef double db;
#define int long long
#define fi first
#define se second
#define mk make_pair
#define pb emplace_back
#define poly vector
#define Bt(a) bitset
#define bc __builtin_popcount
#define pc putchar
#define ci const int&
const int mod = 1e9 + 7;
const db eps = 1e-10;
const int inf = (1 << 30);
inline int Max(ci x, ci y) {return x > y ? x : y;}
inline int Min(ci x, ci y) {return x < y ? x : y;}
inline db Max(db x, db y) {return x - y > eps ? x : y;}
inline db Min(db x, db y) {return x - y < eps ? x : y;}
inline int Add(ci x, ci y, ci M = mod) {return (x + y) % M;}
inline int Mul(ci x, ci y, ci M = mod) {return 1ll * x * y % M;}
inline int Dec(ci x, ci y, ci M = mod) {return (x - y + M) % M;}
typedef pair pii;
inline int Abs(int x) {return x < 0 ? -x : x;}
//char buf[1<<21],*p1=buf,*p2=buf;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char Obuf[105000],*O=Obuf;//Siz shoule be the size of Out File
int pst[30],ptop;
inline void Fprint(){fwrite(Obuf,1,O-Obuf,stdout);}
inline void Fwrite(int x){
if(x==0){*O++='0';return;}
if(x<0)*O++='-',x=-x;ptop=0;
while(x)pst[++ptop]=x%10,x/=10;
while(ptop)*O++=pst[ptop--]+'0';
if(O-Obuf>100000)Fprint(),O=Obuf;
}
inline int read() {
int s = 0, w = 1;
char ch = getchar();
while (!isdigit(ch)) {if (ch == '-') w = -1;ch = getchar();}
while (isdigit(ch)) {s = s * 10 + ch - '0';ch = getchar();}
return s * w;
}
inline void write(int x) {
if (x < 0)putchar('-'), x = -x;
if (x > 9)write(x / 10);
pc(x % 10 + '0');
}
inline int qpow(int x, int y) {
int res = 1;
while (y) {if (y & 1)res = Mul(res, x);x = Mul(x, x);y >>= 1;}
return res;
}
inline void cadd(int &x, int y) {x += y;}
inline void cmul(int &x, int y) {x *= y;}
inline void cmax(int &x, int y) {x = Max(x, y);}
inline void cmin(int &x, int y) {x = Min(x, y);}
const int N = 2e5 + 10;
namespace Refined_heart{
int n,m,k,a[N],b[N];
void solve(){
n=read();m=read();k=read();
for(int i=1;i<=n;++i)a[i]=read();
for(int i=1;i<=m;++i)b[i]=read();
int A=0,B=0;
for(int i=1;i<=n;++i)A+=(m*(k-1)-a[i])%k;
for(int i=1;i<=m;++i)B+=(n*(k-1)-b[i])%k;
if(A%k!=B%k){
puts("-1");
return;
}
int ans=(k-1)*n*m;
ans-=Max(A,B);
cout<