2019中国大学生程序设计竞赛（CCPC） - 网络选拔赛
参考我的博客杜教筛
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1000010;
const ll mod=1e9+7;
int prime[maxn],f[maxn],sum[maxn],cnt;
bool vis[maxn];
int pow_mod(int a,int b) {
int res = 1;
while (b) {
if (b & 1) res = 1ll*res * a % mod;
b >>= 1;
a = 1ll*a * a % mod;
}
return res;
}
void init() {
int N = maxn - 1;
f[1] = 1;
sum[1] = 1;
for (int i = 2; i <= N; i++) {
if (!vis[i]) {
prime[cnt++] = i;
f[i] = i - 1;
}
for (int j = 0; j < cnt && 1ll*prime[j] * i <= N; j++) {
vis[prime[j] * i] = 1;
if (i % prime[j] == 0) {
f[i * prime[j]] = 1ll*f[i] * prime[j]%mod;
break;
}
f[i * prime[j]] = 1ll*f[i] * f[prime[j]]%mod;
}
sum[i]=(sum[i-1]+1ll*f[i]*i%mod)%mod;
}
}
unordered_map<int,int>ma;
int inv6=pow_mod(6,mod-2);
int Go(int n) {
if (n < maxn) return sum[n]%mod;
if (ma[n]!=0) return ma[n];
int ans = 1ll * (n + 1) * n % mod * (2 * n + 1) % mod * inv6 % mod;
for (int i = 2; i <= n; ){
int j = n / (n / i);
ans = (ans + mod - (1ll * (j + 1) * j / 2 - 1ll * i * (i - 1) / 2) % mod * Go(n / i) % mod) % mod;
i = j + 1;
}
return ma[n] = ans;
}
int main() {
init();
int _;
scanf("%d", &_);
int inv2 = pow_mod(2, mod - 2);
while (_--) {
int n,a,b;
scanf("%d%d%d", &n, &a, &b);
printf("%d\n", (Go(n) - 1 + mod) % mod * inv2 % mod);
}
}
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int T;
int n;
ll k;
const int maxn=1e5+5;
priority_queue<ll,vector<ll>,less<ll> >q;
ll c[maxn];
int main()
{
cin>>T;
while(T--)
{
scanf("%d%lld",&n,&k);
while(!q.empty())q.pop();
ll sum=0;
for(register int i=1;i<=n;++i){
scanf("%lld",c+i);
q.push(c[i]);
sum+=c[i];
}
sort(c+1,c+1+n);
ll tot=0;
for(register int i=2;i<=n;++i){
tot+=c[i]/k;
}
if(tot>=n-1){
printf("%lld\n",sum+k);
continue;
}
sum+=n*k;
for(register int i=1;i<n;++i){
ll cur=q.top();
q.pop();
if(cur>k){
sum-=k;
cur-=k;
}
else{
sum-=cur;
cur=0;
}
q.push(cur);
}
printf("%lld\n",sum);
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
char s[12][10000][10000];
void dfs(int id,int cur){
if(id>10)return;
for(register int i=1;i<=cur;++i){
for(register int j=1;j<=cur;++j){
s[id][i][j]=s[id-1][i][j];
}
}
for(register int i=1;i<=cur;++i){
for(register int j=cur+1;j<=2*cur;++j){
s[id][i][j]=s[id-1][i][j-cur];
}
}
for(register int i=cur+1;i<=2*cur;++i){
for(register int j=1;j<=cur;++j){
if(s[id-1][i-cur][j]=='C')s[id][i][j]='P';
else s[id][i][j]='C';
}
}
for(register int i=cur+1;i<=2*cur;++i){
for(register int j=cur+1;j<=2*cur;++j){
s[id][i][j]=s[id-1][i-cur][j-cur];
}
}
dfs(id+1,cur*2);
}
int T,n;
int fast_pow(int a,int b){
int res=1;
while(b){
if(b&1)res=res*a;
a=a*a;
b>>=1;
}
return res;
}
int main() {
s[1][1][1]='C';
s[1][1][2]='C';
s[1][2][1]='P';
s[1][2][2]='C';
dfs(2,2);
scanf("%d",&T);
while(T--){
scanf("%d",&n);
int cur=n;
n=fast_pow(2,n);
//cout<<"***"<<endl;
for(register int i=1;i<=n;++i){
for(register int j=1;j<=n;++j){
printf("%c",s[cur][i][j]);
}
printf("\n");
}
}
return 0;
}
首先询问许多子串的这种问题应该联想到后缀数组，然后我们发现对于后缀数组中的排序（sa[i]表示下标为i的后缀的排名，rk[i]表示排名为i的后缀的下标），每个后缀串与子串的最长公共前缀为该子串的串是连续的，所以我们只要求出这段连续排名的区间（因为在这里的对于子串的最长公共前缀具有单调性，用st表预处理出每个区间的最长公共前缀长度，用二分check即可），然后在这段区间上询问第k大sa[i]的数即可。
我们考虑，一个子串必定是某个后缀的前缀。
排序相邻的后缀他们的前缀一定最相似。
所以全部的一种子串必定是一些排序相邻的后缀的公共前缀。
从l开始的子串，则从rank[l]开始看，两侧height保证大于子串长度，能延伸多长，则证明有多少个这种子串。
我们用ST表维护出height的最小值，然后通过最小值二分即可，边界有些棘手。
然后我们就得到了一个height不小于子串长度的连续区间，这个区间是以原后缀的字典序排序的。
而同时，sa数组下标为排序，值为原串位置。
所以我们对这个区间在sa数组上做主席树，求第k大，即为第k个子串出现的位置。
#include <bits/stdc++.h>
using namespace std;
const int maxn=110010;
int t1[maxn],t2[maxn],c[maxn];
bool cmp(int *r,int a,int b,int l) {
return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(char str[],int sa[],int rank[],int height[],int n,int m) {
int i, j, p, *x = t1, *y = t2;
for (i = 0; i <= m; i++) c[i] = 0;
for (i = 1; i <= n; i++) c[x[i] = str[i]]++;
for (i = 2; i <= m; i++) c[i] += c[i - 1];
for (i = n; i >= 1; i--) sa[c[x[i]]--] = i;
for (j = 1; j <= n; j <<= 1) {
p = 0;
for (i = n - j + 1; i <= n; i++) y[++p] = i;
for (i = 1; i <= n; i++) if (sa[i] > j) y[++p] = sa[i] - j;
for (i = 1; i <= m; i++) c[i] = 0;
for (i = 1; i <= n; i++) c[x[y[i]]]++;
for (i = 2; i <= m; i++) c[i] += c[i - 1];
for (i = n; i >= 1; i--) {
sa[c[x[y[i]]]--] = y[i];
y[i] = 0;
}
swap(x, y);
p = 1;
x[sa[1]] = 1;
for (i = 2; i <= n; i++) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p : ++p;
if (p == n) break;
m = p;
}
int k = 0;
for (i = 1; i <= n; i++) rank[sa[i]] = i;
for (i = 1; i <= n; i++) {
if (rank[i] == 1) continue;
if (k) k--;
j = sa[rank[i] - 1];
while (j + k <= n && str[i + k] == str[j + k]) k++;
height[rank[i]] = k;
}
}
char str[maxn];
int sa[maxn],rk[maxn],height[maxn],n,q,st[maxn][20];
void build_st() {
for (int i = 1; i <= n; i++) st[i][0] = height[i];
for (int k = 1; k <= 19; k++) {
for (int i = 1; i + (1 << k) - 1 <= n; i++) {
st[i][k] = min(st[i][k - 1], st[i + (1 << k - 1)][k - 1]);
}
}
}
int lcp(int a,int b){
a=rk[a];
b=rk[b];
if (a>b) swap(a,b);
if (a==b) return n-sa[a];
int t=log2(b-a);
return min(st[a+1][t],st[b-(1<<t)+1][t]);
}
struct node
{
int l,r,sum;
} hjt[maxn*40];
int cnt,root[maxn*40];
void insert(int l,int r,int pre,int &now,int p)
{
hjt[++cnt]=hjt[pre];
now=cnt;
hjt[now].sum++;
if (l==r)
{
return;
}
int mid=(l+r)>>1;
if (p<=mid)
{
insert(l,mid,hjt[pre].l,hjt[now].l,p);
}
else
{
insert(mid+1,r,hjt[pre].r,hjt[now].r,p);
}
}
int query(int l,int r,int L,int R,int k)
{
if (l==r)
return l;
int mid=(l+r)>>1;
int tmp=hjt[hjt[R].l].sum-hjt[hjt[L].l].sum;
if (k<=tmp)
{
return query(l,mid,hjt[L].l,hjt[R].l,k);
}
else
{
return query(mid+1,r,hjt[L].r,hjt[R].r,k-tmp);
}
}
vector<int>v;
void init(){
cnt=0;
v.clear();
for (int i=0;i<=n;i++) {
root[i]=hjt[i].l=hjt[i].r=hjt[i].sum=0;
}
}
int getid(int x)
{
return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
int main() {
int _;
scanf("%d", &_);
while (_--) {
scanf("%d%d", &n, &q);
scanf("%s", str + 1);
da(str, sa, rk, height, n, 128);
build_st();
init();
for (int i = 1; i <= n; i++) {
v.push_back(sa[i]);
}
sort(v.begin(),v.end());
for (int i=1;i<=n;i++){
insert(1, n, root[i - 1], root[i], getid(sa[i]));
}
while (q--) {
int l, r, k;
scanf("%d%d%d", &l, &r, &k);
int ll, rr, len = r - l + 1;
int left = 1, right = rk[l];
int ans = rk[l];
while (left <= right) {
int mid = (left + right) >> 1;
if (lcp(sa[mid], l) >= len) {
right = mid - 1;
ans = mid;
} else {
left = mid + 1;
}
}
ll = ans;
ans = rk[l];
left = rk[l];
right = n;
while (left <= right) {
int mid = (left + right) >> 1;
if (lcp(sa[mid], l) >= len) {
left = mid + 1;
ans = mid;
} else {
right = mid - 1;
}
}
rr=ans;
if (rr-ll+1<k) printf("-1\n");else{
printf("%d\n",v[query(1,n,root[ll-1],root[rr],k)-1]);
}
}
}
}
不难看出对于每次2操作,答案最大是n+1(因为每次更新是+10000000, 永远不会占用n+1,而且k是保证<=n的). 如果某个数字被进行过1操作, 那么就代表这个数字可以用于2类操作的查询,把这个数字加到set里.对于找2类询问的答案,第一种是从a[r+1,n]里找>=k的最小值(主席树实现),第二种是从这个set里找>=k的最小值(对set进行二分查找),取两者间最小值即可.
#include <bits/stdc++.h>
using namespace std;
const int maxn=200100;
struct node
{
int l,r,sum;
} hjt[maxn*40];
int cnt,root[maxn*20],n,m,book[maxn];
int a[maxn];
set<int>s;
set<int>::iterator it;
void init() {
for (int i = 0; i <= n; i++) book[i] = 0;
s.clear();
root[0] = 0;
cnt = 0;
for (int i = 0; i < maxn * 20; i++) hjt[i].l = hjt[i].r = hjt[i].sum = 0;
}
void insert(int l,int r,int pre,int &now,int p)
{
hjt[++cnt]=hjt[pre];
now=cnt;
hjt[now].sum++;
if (l==r)
{
return;
}
int mid=(l+r)>>1;
if (p<=mid)
{
insert(l,mid,hjt[pre].l,hjt[now].l,p);
}
else
{
insert(mid+1,r,hjt[pre].r,hjt[now].r,p);
}
}
int query(int rt,int l,int r,int L,int R) {
if (R < L) return -1;
if (!rt) return -1;
if (!hjt[rt].sum) return -1;
if (l >= L && r <= R) {
if (l == r)
return l;
else {
int mid = (l + r) >> 1;
int res = -1;
if (L <= mid) res = query(hjt[rt].l, l, mid, L, R);
if (R > mid && res == -1) res = query(hjt[rt].r, mid + 1, r, L, R);
return res;
}
}
int mid = (l + r) >> 1;
int res = -1;
if (L <= mid) res = query(hjt[rt].l, l, mid, L, R);
if (R > mid && res == -1) res = query(hjt[rt].r, mid + 1, r, L, R);
return res;
}
int main() {
int _;
scanf("%d", &_);
while (_--) {
scanf("%d%d", &n, &m);
init();
root[n + 1] = 0;
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = n; i >= 1; i--)
insert(1, n, root[i + 1], root[i], a[i]);
int lastans = 0;
for (int i = 1; i <= m; i++) {
int id;
scanf("%d", &id);
if (id == 1) {
int t;
scanf("%d", &t);
int pos = t ^lastans;
if (book[pos]) continue;
book[pos] = 1;
s.insert(a[pos]);
} else {
int t1, t2;
scanf("%d%d", &t1, &t2);
int R = (t1 ^lastans) + 1;
int k = t2 ^lastans;
int ans = n + 1;
it = s.lower_bound(k);
if (it != s.end()) ans = *it;
if (R <= n) {
t1 = query(root[R], 1, n, k, n);
if (t1 != -1) ans = min(ans, t1);
}
printf("%d\n", ans);
lastans = ans;
}
}
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a,b;
int main()
{
int T;
cin>>T;
while(T--)
{
cin>>a>>b;
ll ans=(a&b);
if(ans==0){
ans=1;
}
printf("%lld\n",ans);
}
return 0
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=5e4+10;
ll n,m,q,Q[maxn],ans[maxn],kk;
typedef pair<ll,int> PII;
vector<PII>G[maxn];
multiset<PII>s;
multiset<PII>::iterator it,it1;
int main() {
int _;
scanf("%d", &_);
while (_--) {
s.clear();
kk=0;
scanf("%lld%lld%lld", &n, &m, &q);
for (int i = 1; i <= n; i++) G[i].clear();
for (int i = 1, u, v, w; i <= m; i++) {
scanf("%d%d%d",&u,&v,&w);
G[u].push_back(make_pair(w, v));
s.insert(make_pair(w, v));
}
for (int i = 1; i <= n; i++) {
sort(G[i].begin(), G[i].end());
}
for (int i = 1; i <= q; i++) {
scanf("%lld", &Q[i]);
kk = max(kk, Q[i]);
}
for (int i = 1; i <= kk; i++) {
it = s.begin();
it1 = s.end();
it1--;
ans[i] = it->first;
int sz1 = s.size();
int sz = G[it->second].size();
for (int j = 0; j < sz; j++) {
if (s.size() >= kk-i+1 && it->first + G[it->second][j].first >= it1->first) break;
s.insert(make_pair(it->first + G[it->second][j].first, G[it->second][j].second));
}
s.erase(it);
while (s.size()>=kk-i+1) it=s.end(),--it,s.erase(it);
}
for (int i = 1; i <= q; i++) printf("%lld\n", ans[Q[i]]);
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int n,m;
deque<int>q;
int a[maxn],ans[maxn],vis[maxn],x,k;
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = 1; i <= m; i++) {
scanf("%d", &x);
q.push_front(x);
}
while (!q.empty()) {
int x = q.front();
if (vis[x] == 0) ans[k++] = x;
q.pop_front();
vis[x] = 1;
}
for (int i = 1; i <= n; i++) {
if (vis[i] == 0) {
ans[k++] = i;
}
}
for (int i = 0; i < k; i++) {
printf("%d ", ans[i]);
}
//printf("\n");
}
