充实的感情产生出了希冀 机械的手心不自主向人类学习 你想得到什么东西
test18
签到题sign
好像做过，设 \(f(i)\) 表示选定 \(k=i\) 时的答案，这个显然可以 \(O(n)\) 求出来，写一个 nth_element 即可。
首先 \(f(n-k+1)\) 最大，具体的可以假设最后一个区间被选取，因为区间包含，贡献可以一直替换到一个不选的位置，最小长度二分就好了。yy 一下好像也可以查找每一个位置的最小贡献长度然后直接算出来。
#define _CRT_SECURE_NO_WARNINGS
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include<bits/stdc++.h>
#define int long long
#define up(i,l,r) for(int i=l; i<=r; ++i)
#define dn(i,r,l) for(int i=r; i>=l; --i)
using namespace std;
const int N=100005, M=21;
int T, n, k, t, len, m, a[N], mx[N][M], mi[N][M], lg[N], val[N];
int getmax(int l,int r) {
int k=lg[r-l+1];
return max(mx[l][k],mx[r-(1<<k)+1][k]);
}
int getmin(int l,int r) {
int k=lg[r-l+1];
return min(mi[l][k],mi[r-(1<<k)+1][k]);
}
int solve(int l) {
m=0;
up(i,1,n-l+1) val[++m]=getmax(i,i+l-1)-getmin(i,i+l-1);
nth_element(val+1,val+(m-k+1),val+1+m);
return val[m-k+1];
}
void mian() {
cin >> n >> k, t=log2(n), len=n-k+1;
up(i,1,n) cin >> a[i], mx[i][0]=mi[i][0]=a[i], lg[i]=log2(i);
up(j,1,t) up(i,1,n-(1<<j)+1) {
mx[i][j]=max(mx[i][j-1],mx[i+(1<<j-1)][j-1]);
mi[i][j]=min(mi[i][j-1],mi[i+(1<<j-1)][j-1]);
}
int l=1, r=n-k+1, p, ans=solve(len);
while(l<=r) {
int mid=(l+r)>>1;
if(solve(mid)==ans) p=mid, r=mid-1;
else l=mid+1;
}
cout << ans << ' ' << p << '\n';
}
signed main() {
freopen("sign.in","r",stdin);
freopen("sign.out","w",stdout);
ios::sync_with_stdio(0);
cin.tie(0);
cin >> T;
while(T--) mian();
return 0;
}
爬山游戏climb
对于每一个 \(x\)，考虑存不存在必胜的 \(y\)，分类讨论即可，相互干扰的地方需要讨论一下奇偶性，懒得赘述惹。
#include<bits/stdc++.h>
#define int long long
#define up(i,l,r) for(int i=l; i<=r; ++i)
#define dn(i,r,l) for(int i=r; i>=l; --i)
#define pb push_back
using namespace std;
const int N=100005;
int T, n, a[N], lwl[N], lwr[N], upl[N], upr[N], pre[N], suf[N];
vector<int> ans;
void mian() {
ans.clear();
cin >> n;
up(i,1,n) cin >> a[i];
up(i,1,n) {
lwl[i]=(i>1&&a[i-1]<a[i])?lwl[i-1]:i;
upl[i]=(i>1&&a[i-1]>a[i])?upl[i-1]:i;
}
dn(i,n,1) {
lwr[i]=(i<n&&a[i+1]<a[i])?lwr[i+1]:i;
upr[i]=(i<n&&a[i+1]>a[i])?upr[i+1]:i;
}
up(i,1,n) pre[i]=suf[i]=max(i-upl[i],upr[i]-i);
up(i,1,n) pre[i]=max(pre[i-1],pre[i]);
dn(i,n,1) suf[i]=max(suf[i+1],suf[i]);
up(i,1,n) {
int l=lwl[i], r=lwr[i];
bool flag=max(pre[l-1],suf[r+1])>=max(i-l,r-i);
if((i-l+1)%2==0) flag|=max(l-upl[l],upr[l]-l)>=(r-i);
else {
flag|=(i-l<=l-lwl[l])&&(r-i<=max(i-l,l-upl[l]));
flag|=(i-l-1>=r-i);
}
if((r-i+1)%2==0) flag|=max(r-upl[r],upr[r]-r)>=(i-l);
else {
flag|=(r-i<=upr[r]-r)&&(i-l<=max(r-i,upr[r]-r));
flag|=(r-i-1>=i-l);
}
if(!flag) ans.pb(i);
}
cout << ans.size() << ' ';
for(int i:ans) cout << i << ' ';
cout << '\n';
}
signed main() {
freopen("climb.in","r",stdin);
freopen("climb.out","w",stdout);
ios::sync_with_stdio(0);
cin.tie(0);
cin >> T;
while(T--) mian();
return 0;
}
旅行travel
为了方便不妨先考虑 \(k=1\) 的点集 \(p=\{x_i,y_i\}\) 还有总位移，然后会出现的点就是 \((x_i+t\Delta x,y_i+t\Delta y)\)。
我们对于 \(i\) 考虑 \((x_i+t\Delta x,y_i+t\Delta y)\) 的合法性。首先要跟别的 \(i’\) 去重，重复的点出现的形式显然是 \(l/r\) 秒前/后的 \(i'\) 会出现，比如 \((x_i,y_i)=(x_j+r\Delta x,y_j+r\Delta y)\)，要求 \(r=\frac{x_i-x_j}{\Delta x}=\frac{y_i-y_j}{\Delta y}\)，按照 \(x\mod \Delta x,x\Delta y-y\Delta x\) 分类同类里面有贡献，找 \(l/r\) 就相当于找同类更小/大的 \(x_{i'}\)。不妨钦定 \(i\) 是第一个出现的，那么要求未来不会出现，就是 \(t=k-\max r,\dots k-1\)。之后考虑 \((x'(+1),y'(+1))\) 的存在性，还是先查询一个 \(l/r\) 秒前/后会出现，合法的 \(t\) 至少满足 \(t\geq l\) 或者 \(t\leq k-1-r\)。做一个区间取并即可，想要好些一点可以离散化区间起点然后暴力判断。
#include<bits/stdc++.h>
#define int long long
#define up(i,l,r) for(int i=l; i<=r; ++i)
#define dn(i,r,l) for(int i=r; i>=l; --i)
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define x first
#define y second
using namespace std;
const int N=100005, inf=2e9;
int T, n, k, X, Y, Ans, sp[N], tot;
pii a[N];
char str[N];
map<pii,set<int> > qur;
pii get(int x,int y) {
pii u=mp((x%X+X)%X,x*Y-y*X);
if(Y==0) u.second=y;
auto L=qur[u].upper_bound(x);
auto R=qur[u].lower_bound(x);
pii res=mp(inf,inf);
if(L!=qur[u].begin()) res.second=(x-*--L)/abs(X);
if(R!=qur[u].end()) res.first=(*R-x)/abs(X);
return res;
}
void add(int x,int y,int v) {
pii u=mp((x%X+X)%X,x*Y-y*X);
if(Y==0) u.second=y;
if(v==1) qur[u].insert(x); else qur[u].erase(x);
}
void mian() {
qur.clear(), Ans=0;
cin >> n >> k >> (str+1);
up(i,1,n) {
a[i]=a[i-1];
if(str[i]=='E') ++a[i].x;
if(str[i]=='N') ++a[i].y;
if(str[i]=='W') --a[i].x;
if(str[i]=='S') --a[i].y;
}
X=a[n].x, Y=a[n].y, a[++n]=mp(0,0);
sort(a+1,a+1+n), n=unique(a+1,a+1+n)-a-1;
if(X<0) { X=-X; up(i,1,n) a[i].x=-a[i].x; }
if(X==0) { swap(X,Y); up(i,1,n) swap(a[i].x,a[i].y); }
if(X==0) {
map<pii,int> chk;
up(i,1,n) chk[a[i]]=1;
up(i,1,n) if(chk[mp(a[i].x+1,a[i].y)]&&chk[mp(a[i].x,a[i].y+1)]&&chk[mp(a[i].x+1,a[i].y+1)]) ++Ans;
cout << Ans << '\n';
return;
}
up(i,1,n) add(a[i].x,a[i].y,1);
up(i,1,n) {
add(a[i].x,a[i].y,-1);
int L=get(a[i].x,a[i].y).second, R=k-1;
if(L==inf||L>k) L=0; else L=k-L;
add(a[i].x,a[i].y,+1);
pii x=get(a[i].x+1,a[i].y), y=get(a[i].x,a[i].y+1), z=get(a[i].x+1,a[i].y+1);
sp[tot=0]=0, sp[++tot]=L, sp[++tot]=R+1;
sp[++tot]=x.x, sp[++tot]=R-x.y+1;
sp[++tot]=y.x, sp[++tot]=R-y.y+1;
sp[++tot]=z.x, sp[++tot]=R-z.y+1;
sort(sp+1,sp+1+tot), tot=unique(sp+1,sp+1+tot)-sp-1;
up(i,1,tot-1) {
if(sp[i]<L||sp[i+1]-1>R) continue;
int u=sp[i];
if((x.x<=u||x.y<=R-u)&&(y.x<=u||y.y<=R-u)&&(z.x<=u||z.y<=R-u)) Ans+=sp[i+1]-sp[i];
}
}
cout << Ans << '\n';
}
signed main() {
freopen("travel.in","r",stdin);
freopen("travel.out","w",stdout);
ios::sync_with_stdio(0);
cin.tie(0);
cin >> T;
while(T--) mian();
return 0;
}
排列permutation
先组合意义一下，一组 \(p\) 的贡献相当于 \(a_i\leq p_{j,i}\) 的 \(a\) 的数量，所以想办法 dp 一下对于 \(a\) 有多少组合法的 \(p\)，答案就是所有 \(a\) 的计数加起来。考虑一下 \(a\) 确定的时候怎么做，填写一个 \(p_{i,j}\) 的时候关心 \(a_i\)，那么初始有 \(n-a_i+1\) 种填法，那还关心以前填了多少个 \(\geq a_i\) 的数量。非常经典又显然的可以考虑按照 \(a\downarrow\) 填写，填法就是 \(n-a_i+1-pre\)。
放到序列上看看怎么做，关心哪些列填写了以及现在要填写的值，状压所有不太实际，但是因为不用钦定的列等价可以 \(2^{10}\) 状压哪些要被钦定的列被填写了、 再记录有多少不被钦定的列被填写了。转移的话先虑多填写几列不被钦定的列，然后还要同层转移钦定的列的填写，同层顺序转移即可。
诶呀怎么说的这么少，有的人写的时候可不是这样的轻松 /fad
#define _CRT_SECURE_NO_WARNINGS
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include<bits/stdc++.h>
#define int long long
#define up(i,l,r) for(int i=l; i<=r; ++i)
#define dn(i,r,l) for(int i=r; i>=l; --i)
using namespace std;
const int N=55, M=11, P=998244353;
int id, T, n, m, k, x[M], y[M], xt, len, yh[N][N], tag[N][N], sum[N][N];
int range[N][1<<M][N], Range[N][1<<M][N], sigma[N][1<<M][N];
int rx[M], ry[M], w[M], f[N][N][1<<M], popcnt[1<<M], msk[N], inv[N], coef[N][N][N][M];
inline void add(int &a,int b) { a=(a+b)%P; }
inline void mul(int &a,int b) { a=a*b%P; }
inline int C(int n,int m) { return (m<0||n<m)?0:yh[n][m]; }
inline int ksm(int a,int b=P-2) {
int ret=1;
for( ; b; b>>=1) {
if(b&1) ret=ret*a%P;
a=a*a%P;
}
return ret;
}
void mian() {
xt=len=0;
memset(tag,0,sizeof(tag));
memset(sum,0,sizeof(sum));
memset(f,0,sizeof(f));
memset(popcnt,0,sizeof(popcnt));
memset(msk,0,sizeof(msk));
cin >> n >> m >> k;
up(i,1,k) {
cin >> rx[i] >> ry[i] >> w[i];
x[++xt]=rx[i], y[++len]=ry[i];
}
sort(x+1,x+1+xt), xt=unique(x+1,x+1+xt)-x-1;
sort(y+1,y+1+len), len=unique(y+1,y+1+len)-y-1;
up(i,1,n) inv[i]=ksm(i,m-xt);
up(i,1,k) {
rx[i]=lower_bound(x+1,x+1+xt,rx[i])-x;
ry[i]=lower_bound(y+1,y+1+len,ry[i])-y;
tag[rx[i]][ry[i]]=w[i], sum[rx[i]][w[i]]=1, msk[rx[i]]|=(1<<ry[i]-1);
}
up(i,1,k) up(j,1,i-1) if(w[i]==w[j]&&rx[i]==rx[j]) { cout << 0 << '\n'; return; }
up(i,1,xt) dn(j,n,1) sum[i][j]=sum[i][j+1]+1-sum[i][j];
up(i,1,(1<<len)-1) popcnt[i]=popcnt[i&(i-1)]+1;
f[0][0][0]=1;
up(i,1,n) up(j,0,n-len) up(d,0,n-len-j) up(ps,0,len) {
coef[i][j][d][ps]=1;
up(u,j+1,j+d) mul(coef[i][j][d][ps],inv[max(0ll,(n-i+1)-(n-ps-u))]);
}
up(i,1,n) up(s,0,(1<<len)-1) {
range[i][s][0]=Range[i][s][0]=1;
up(u,1,n-len) {
range[i][s][u]=range[i][s][u-1];
sigma[i][s][u]=sigma[i][s][u-1];
up(p,1,xt) {
int val=max(0ll,sum[p][i]-(sum[p][1]-u-popcnt[s]+popcnt[msk[p]&s]));
mul(range[i][s][u],max(val,1ll));
add(sigma[i][s][u],val==0);
}
Range[i][s][u]=ksm(range[i][s][u]);
}
}
up(i,1,n) {
up(j,0,n-len) up(s,0,(1<<len)-1) up(d,0,n-len-j) {
int val=C(n-len-j,d)*coef[i][j][d][popcnt[s]]%P;
if(sigma[i][s][j+d]-sigma[i][s][j]) val=0;
else (val*=range[i][s][j+d]*Range[i][s][j]%P)%=P,
add(f[i][j+d][s],f[i-1][j][s]*val%P);
}
up(u,1,len) up(j,0,n-len) up(s,0,(1<<len)-1) {
if(s>>(u-1)&1) continue;
int val=inv[max(0ll,(n-i+1)-(n-j-popcnt[s]-1))];
up(p,1,xt) {
if(tag[p][u]) val*=(tag[p][u]>=i);
else mul(val,max(0ll,sum[p][i]-(sum[p][1]-j-1-popcnt[s]+popcnt[msk[p]&s])));
}
add(f[i][j][s+(1<<u-1)],f[i][j][s]*val%P);
}
}
cout << f[n][n-len][(1<<len)-1] << '\n';
}
signed main() {
freopen("permutation.in","r",stdin);
freopen("permutation.out","w",stdout);
ios::sync_with_stdio(0);
cin.tie(0);
up(i,0,50) {
yh[i][0]=yh[i][i]=1;
up(j,1,i-1) yh[i][j]=(yh[i-1][j-1]+yh[i-1][j])%P;
}
cin >> id >> T;
while(T--) mian();
return 0;
}
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