2021“MINIEYE杯”中国大学生算法设计超级联赛1
2021“MINIEYE杯”中国大学生算法设计超级联赛1
1001 Mod, Or and Everything
题目大意
求 \(\sum_{i=1}^{n-1}n\ mod\ i\)
\(n<=10^{12}\)
打表找规律n的答案是小于n的第一个\(2^x\)再减去1。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define int long long
int read(){
int sum=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
return sum*f;
}
int work(int x){
int num=0;
x>>=1;
while(x){
x>>=1;
num=num<<1|1;
}
return num;
}
signed main(){
int T=read();
while(T--){
int ans=0;
int n=read();
cout<<work(n-1)<<endl;
}
return 0;
}
1002 Rocket land
题解说是K-Dtree裸题，APIO2018D1T1？
但是我的K-Dtree T了？
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
#define mid (L+R>>1)
const int N=501000;
const int K=2;
const double e=0.75;
int sum[N],ch[N][2],val[N],mx[N][K],mn[N][K],d[N][K],size[N];
int now_D,stack[N],top,tot,root;
int a[N],b[2][K];
int read(){
int sum=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
return sum*f;
}
void update(int x){
int l=ch[x][0],r=ch[x][1];
sum[x]=sum[l]+sum[r]+val[x];
size[x]=size[l]+size[r]+1;
if(l)mx[x][0]=max(mx[x][0],mx[l][0]),mn[x][0]=min(mn[x][0],mn[l][0]);
if(r)mx[x][0]=max(mx[x][0],mx[r][0]),mn[x][0]=min(mn[x][0],mn[r][0]);
if(l)mx[x][1]=max(mx[x][1],mx[l][1]),mn[x][1]=min(mn[x][1],mn[l][1]);
if(r)mx[x][1]=max(mx[x][1],mx[r][1]),mn[x][1]=min(mn[x][1],mn[r][1]);
}
bool cmp(int x,int y){
return d[x][now_D]<d[y][now_D];
}
int build(int L,int R,int D){
if(L>R)return 0;
now_D=D;
nth_element(stack+L,stack+mid,stack+R+1,cmp);
int now=stack[mid];
sum[now]=val[now];
mx[now][0]=mn[now][0]=d[now][0];
mx[now][1]=mn[now][1]=d[now][1];
ch[now][0]=build(L,mid-1,(D+1)%K);
ch[now][1]=build(mid+1,R,(D+1)%K);
update(now);
return now;
}
void recycle(int now){
if(ch[now][0])recycle(ch[now][0]);
stack[++top]=now;
if(ch[now][1])recycle(ch[now][1]);
}
void rebuild(int &now,int D){
top=0;recycle(now);
now=build(1,top,D);
}
bool check(int now){
return size[ch[now][0]]>e*size[now]||size[ch[now][1]]>e*size[now];
}
void ins(int &x,int w,int D){
if(!x){
x=++tot;
size[x]=1;sum[x]=val[x]=w;
ch[x][0]=ch[x][1]=0;
mx[x][0]=mn[x][0]=d[x][0]=a[0];
mx[x][1]=mn[x][1]=d[x][1]=a[1];
return;
}
if(a[D]<d[x][D])ins(ch[x][0],w,(D+1)%K);
else ins(ch[x][1],w,(D+1)%K);
update(x);
if(check(x))rebuild(x,D);
}
bool judge_all(int x,int y,int r,int now){
if(!now)return false;
LL tmp=0,cnt=0;
if(1ll*(mn[now][0]-x)*(mn[now][0]-x)+1ll*(mn[now][1]-y)*(mn[now][1]-y)<=1ll*r*r)cnt++;
if(1ll*(mx[now][0]-x)*(mx[now][0]-x)+1ll*(mn[now][1]-y)*(mn[now][1]-y)<=1ll*r*r)cnt++;
if(1ll*(mn[now][0]-x)*(mn[now][0]-x)+1ll*(mx[now][1]-y)*(mx[now][1]-y)<=1ll*r*r)cnt++;
if(1ll*(mx[now][0]-x)*(mx[now][0]-x)+1ll*(mx[now][1]-y)*(mx[now][1]-y)<=1ll*r*r)cnt++;
if(cnt==4)return true;
return false;
}
bool judge_node(int x,int y,int r,int now){
if(!now)return false;
LL tmp=1ll*(d[now][0]-x)*(d[now][0]-x)+1ll*(d[now][1]-y)*(d[now][1]-y);
if(tmp<=1ll*r*r)return true;
else return false;
}
bool judge(int x,int y,int r,int now){
if(!now)return false;
int x_1=x-r,x_2=x+r,y_1=y-r,y_2=y+r;
int x_mid=(x_1+x_2),y_mid=(y_1+y_2),X_mid=(mn[now][0]+mx[now][0]),Y_mid=(mn[now][1]+mx[now][1]);
if(1LL*abs(x_mid-X_mid)<=1LL*x_2-x_1+mx[now][0]-mn[now][0]&&1LL*abs(y_mid-Y_mid)<=1LL*y_2-y_1+mx[now][1]-mn[now][1])return true;
else return false;
}
LL check(int x,int y,int r,int now){
LL ans=0;
if(judge_all(x,y,r,now))return sum[now];
if(!judge(x,y,r,now))return 0;
if(judge_node(x,y,r,now))ans+=val[now];
if(judge(x,y,r,ch[now][0]))ans+=check(x,y,r,ch[now][0]);
if(judge(x,y,r,ch[now][1]))ans+=check(x,y,r,ch[now][1]);
return ans;
}
int main(){
// freopen("1.in","r",stdin);
// freopen("xdx.out","w",stdout);
int T=read();
while(T--){
int n=read();
for(int i=1;i<=n;i++){
for(int j=0;j<K;j++)a[j]=read();
int w=read();
ins(root,w,0);
printf("%lld\n",check(a[0],a[1],read(),root));
}
tot=root=0;
}
return 0;
}
1005 Minimum spanning tree
求2~n组成边权为\(lcm(u[i],v[i])\)生成树的边权之和的最小值。
为使答案最小，质数连2，合数连一个约数。统计答案即可。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define int long long
#define MAXN 1e7
#define N 10000000
int read(){
int sum=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
return sum*f;
}
int vis[N],phi[N],pri[N],cnt,ans[N];
void init() {
phi[1] = 1;//0是质数
for (int i = 2; i < MAXN; ++i) {
if (!vis[i]) {
phi[i] = i - 1;
pri[cnt++] = i;
}
for (int j = 0; j < cnt; ++j) {
if (1ll * i * pri[j] >= MAXN) break;
vis[i * pri[j]] = 1;
if (i % pri[j]) {
phi[i * pri[j]] = phi[i] * (pri[j] - 1);
} else {
phi[i * pri[j]] = phi[i] * pri[j];
break;
}
}
}
}
void pre_work(){
ans[2]=0;
for(int i=3;i<=N;i++){
if(vis[i]==0)ans[i]=ans[i-1]+i*2;
else ans[i]=ans[i-1]+i;
}
}
signed main(){
int T=read();
init();
pre_work();
while(T--){
int n=read();
cout<<ans[n]<<endl;
}
}
1006 Xor sum
给一个数列，求一个最短的区间使其区间异或和大于k
\(n<=100000\)
01trie上维护子树中出现过的前缀异或和在数列中最大位置。然后对于每一个r判断从1~r找到最优答案。
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
#define INF 2147483647
#define N 500100
int ch[N*50][2],mx_pos[N*50],root,tot,sum[N];
int read(){
int sum=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
return sum*f;
}
void update(int now){
if(ch[now][0])mx_pos[now]=max(mx_pos[ch[now][0]],mx_pos[now]);
if(ch[now][1])mx_pos[now]=max(mx_pos[ch[now][1]],mx_pos[now]);
}
void add(int tall,int pos,int x,int &now){
if(now==0)now=++tot;
if(tall==0){mx_pos[now]=pos;return;}
int bit=x>>tall-1&1;
add(tall-1,pos,x,ch[now][bit]);
update(now);
}
int check(int tall,int x,int k,int now){
int bit_k=k>>tall-1&1;
int bit_x=x>>tall-1&1;
if(tall==0)return mx_pos[now];
if(bit_k==0)return max(check(tall-1,x,k,ch[now][bit_x]),mx_pos[ch[now][bit_x^1]]);
else return mx_pos[ch[now][bit_x^1]]==-1?-1:check(tall-1,x,k,ch[now][bit_x^1]);
}
void init(){
while(tot){
mx_pos[tot]=-1;
ch[tot][0]=ch[tot][1]=0;
tot--;
}
root=0;
}
int main(){
int T=read();
memset(mx_pos,-1,sizeof(mx_pos));
while(T--){
init();
int ans=INF,ansl=INF,ansr=INF;
int n=read(),k=read();
add(31,0,0,root);
for(int i=1;i<=n;i++){
sum[i]=sum[i-1]^read();
int l=check(31,sum[i],k,root)+1;
add(31,i,sum[i],root);
if(l==0)continue;
int tmp=i-l+1;
if(tmp<ans)ans=tmp,ansl=l,ansr=i;
}
if(ans==INF)printf("-1\n");
else printf("%d %d\n",ansl,ansr);
}
return 0;
}
1007 Pass!
打表然后发现\(f[i]=f[i-1]*(n-1)+(-1)^x(n-1)\)
推一推式子得到\(\frac{n-1}{n}[(n-1)^{t-1}+(-1)^t]=x\)
然后对于t的奇偶讨论，用BSGS求答案就好。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<unordered_map>
using namespace std;
#define N 40000
#define int long long
unordered_map<int,int> Map1,Map2;
int p,m;
int ksm(int x,int b){
int ans=1;
while(b){
if(b&1)ans=ans*x%p;
b>>=1;
x=x*x%p;
}
return ans;
}
int BSGS(int n,int x1,int x2){
int tmp=1;
for(int i=1;i<=m;i++){
tmp=tmp*n%p;
Map1[tmp*x1%p]=i;
Map2[tmp*x2%p]=i;
}
n=ksm(n,m);tmp=1;
for(int i=1;i<=m+1;i++){
tmp=tmp*n%p;
int w1=Map1.count(tmp);
int w2=Map2.count(tmp);
if(w2)return (i*m-Map2[tmp]+p)*2%p;
if(w1)return ((i*m-Map1[tmp]+p)*2+1)%p;
}
return -1;
}
int read(){
int sum=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
return sum*f;
}
signed main(){
p=998244353;m=sqrt(p)+1;
int T=read();
while(T--){
int n=read(),x=read();
if(x==1){
printf("0\n");
continue;
}
if(x==0){
printf("1\n");
continue;
}
x=x*n%p*ksm(n-1,p-2)%p;
int base=(n-1)*(n-1)%p;
Map1.clear();
Map2.clear();
int f1=(x+1)%p;
int f2=(x-1+p)*(n-1)%p;
int ans=BSGS(base,f1,f2);
printf("%lld\n",ans);
}
return 0;
}
1008 Maximal submatrix
洛谷原题，《月蟾宫》
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m,s[2010],l[2010],ans,a[2010][2010],num[2010][2010];
int main(){
int T;
scanf("%d",&T);
while(T--){
int n,m;
ans=0;
memset(a,0,sizeof(a));
memset(num,0,sizeof(num));
memset(s,0,sizeof(s));
memset(l,0,sizeof(l));
scanf("%d %d",&n,&m);
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j){
scanf("%d",&num[i][j]);
if(num[i][j]>=num[i-1][j]) a[i][j]=a[i-1][j]+1;
else a[i][j]=1;
}
for(int top,len,i=1;i<=n;++i){
top=0; len=0; s[top]=0; l[top]=0;
for(int j=1;j<=m;++j){
if(a[i][j]>=s[top]){
s[++top]=a[i][j];
l[top]=1;
}else{
len=0;
while(top&&s[top]>a[i][j]){
len+=l[top];
ans=max(ans,len*s[top]);
--top;
}
s[++top]=a[i][j];
l[top]=len+1;
}
}
len=0;
while(top){
len+=l[top];
ans=max(ans,len*s[top]);
--top;
}
}
printf("%d\n",ans);
}
return 0;
}
1009 KD-Graph
二分权值用并查集维护连通块即可
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define N 101000
#define M 501000
int fa[N],x[M],w[M],y[M],col[N],n,m,k;
int find(int x){
if(fa[x]==x)return x;
else return fa[x]=find(fa[x]);
}
int check(int W){
for(int i=1;i<=n;i++)fa[i]=i;
for(int i=1;i<=m;i++){
if(w[i]<=W){
int fx=find(x[i]);
int fy=find(y[i]);
fa[fx]=fy;
}
}
for(int i=1;i<=n;i++)col[i]=0;
for(int i=1;i<=n;i++)col[find(i)]=1;
int ans=0;
for(int i=1;i<=n;i++)ans+=col[i];
return ans;
}
int read(){
int sum=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
return sum*f;
}
int main(){
int T=read();
while(T--){
n=read(),m=read(),k=read();
for(int i=1;i<=m;i++)x[i]=read(),y[i]=read(),w[i]=read();
int l=0,r=1000000001;
int ans=-1;
while(l<=r){
int mid=(l+r)>>1;
int tmp=check(mid);
if(tmp<=k){
if(tmp==k)ans=mid;
r=mid-1;
}
else l=mid+1;
}
printf("%d\n",ans);
}
return 0;
}
1010 zoto
HH项链原题，就是询问一个区间的某个值域有多少不同的数。
离线然后树状数组套权值线段树。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define N 101000
#define mid (L+R>>1)
int sum[N*400],ch[N*400][2],tot,n,m,root[N*20];
int a[N],pre[N],pos[N],ans[N];
struct qu{
int x0,x1,y0,y1,num;
}q[N];
int read(){
int sum=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
return sum*f;
}
void update(int now){
sum[now]=sum[ch[now][0]]+sum[ch[now][1]];
}
void add(int L,int R,int x,int w,int &now){
if(now==0)now=++tot;
if(L==R){
sum[now]+=w;
return;
}
if(x>mid)add(mid+1,R,x,w,ch[now][1]);
else add(L,mid,x,w,ch[now][0]);
update(now);
}
int check(int L,int R,int l,int r,int now){
if(now==0)return 0;
if(L==l&&R==r)return sum[now];
if(l>mid)return check(mid+1,R,l,r,ch[now][1]);
else if(r<=mid)return check(L,mid,l,r,ch[now][0]);
else return check(L,mid,l,mid,ch[now][0])+check(mid+1,R,mid+1,r,ch[now][1]);
}
bool cmp(qu a,qu b){
return a.x1<b.x1;
}
int lowbit(int x){
return x&-x;
}
void ADD(int x,int val,int w){
for(int i=x;i<=n;i+=lowbit(i)){
add(1,100001,val,w,root[i]);
}
}
int query(int x,int l,int r){
int ans=0;
for(int i=x;i>=1;i-=lowbit(i)){
ans+=check(1,100001,l,r,root[i]);
}
return ans;
}
int main(){
int T=read();
while(T--){
while(tot){
ch[tot][0]=ch[tot][1]=sum[tot]=0;
tot--;
}
n=m=0;
memset(ans,0,sizeof(ans));
memset(pos,0,sizeof(ans));
memset(pre,0,sizeof(ans));
memset(a,0,sizeof(ans));
memset(root,0,sizeof(ans));
n=read(),m=read();
for(int i=1;i<=n;i++){
a[i]=read()+1;
pre[i]=pos[a[i]]+1;
pos[a[i]]=i;
}
for(int j=1;j<=m;j++)q[j].x0=read(),q[j].y0=read()+1,q[j].x1=read(),q[j].y1=read()+1,q[j].num=j;
sort(q+1,q+1+m,cmp);
for(int now=0,i=1;i<=m;i++){
while(now<q[i].x1){
now++;
ADD(pre[now],a[now],+1);
ADD(now+1,a[now],-1);
}
ans[q[i].num]=query(q[i].x0,q[i].y0,q[i].y1);
}
for(int i=1;i<=m;i++)
printf("%d\n",ans[i]);
}
return 0;
}
