（有任何问题欢迎留言或私聊 && 欢迎交流讨论哦
组队训练赛第一次AK，emm，贴部分代码和简单题解吧
目录
A题Zero Array:
离散化+数组模拟
AC代码：
#include<bits/stdc++.h>
#define mme(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const int N = 1e6 + 7;
const int ME = 1e5 + 7;
const int mod = 998244353;
const int INF = 0x3f3f3f3f;
int n, m, q;
struct lp{
int opt,x,y;
}op[ME];
int vis[N],ar[N],br[N],now[N];
int main(){
int tim;
scanf("%d",&tim);
while(tim--){
scanf("%d%d", &n, &q);
mme(vis,0);int tot=0;
for(int i=1;i<=n;++i){
scanf("%d",&ar[i]);
br[tot++]=ar[i];
}
br[tot++]=0;
for(int i=0;i<q;++i){
int opt,x,y;
scanf("%d",&opt);
op[i].opt=opt;
if(opt==1){
scanf("%d%d",&x,&y);
op[i].x=x;op[i].y=y;
br[tot++]=y;
}
}
sort(br,br+tot);
int k = unique(br,br+tot)-br,sum=0;
for(int i=1;i<=n;++i){
now[i]=lower_bound(br,br+k,ar[i])-br+1;
if(now[i]==1)continue;
vis[now[i]]++;
if(vis[now[i]]==1)sum++;
}
for(int i=0;i<q;++i){
if(op[i].opt==2){
printf("%d\n", sum);
}else{
int x=op[i].x,y=op[i].y;
if(now[x]!=1){
vis[now[x]]--;
if(vis[now[x]]==0)sum--;
}
now[x]=lower_bound(br,br+k,y)-br+1;
if(now[x]!=1){
vis[now[x]]++;
if(vis[now[x]]==1)sum++;
}
}
}
}
return 0;
}B题New Assignment:
素数筛+二分图最大匹配
AC代码
#include<bits/stdc++.h>
#define mme(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const int N = 2e6 + 7;
const int ME = 1e6 + 7;
const int M = 1e4 + 7;
const int mod = 998244353;
const int INF = 0x3f3f3f3f;
int n, m, top;
int noprime[N], pcnt, p[N/2];
vector<int> have[N];
int ar[M];
bool is[M];
struct Hopcroft_Carp{
static const int N = 20000 + 5;
static const int M = 150000 + 5;
static const int oo = 0x3f3f3f3f;
int n1,n2,res,tot;
int ddx[N],ddy[N],mx[N],my[N];
int que[N<<1];
bool vis[N<<1];
int head[N<<1];
int nex[M],u[M],v[M];
void clear(){
tot=-1;
res=0;
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
}
void add(int x,int y){
u[++tot]=x;v[tot]=y;
nex[tot]=head[x];
head[x]=tot;
}
bool bfs(){
memset(ddx,-1,sizeof(ddx));
memset(ddy,-1,sizeof(ddy));
res=oo;
int rear=1,tail=1;
for(int i=1;i<=n1;++i){
if(mx[i]==-1){
que[++tail]=i;ddx[i]=0;
}
}
while(rear<=tail){
int x=que[rear];
if(ddx[x]>res)break;
for(int i=head[x];~i;i=nex[i]){
if(ddy[v[i]]!=-1)continue;
ddy[v[i]]=ddx[x]+1;
if(my[v[i]]==-1)res=ddy[v[i]];
else {
ddx[my[v[i]]]=ddy[v[i]]+1;
que[++tail]=my[v[i]];
}
}
rear++;
}
return res!=oo;
}
bool dfs(int x){
for(int i=head[x];~i;i=nex[i]){
if(vis[v[i]]==0&&ddy[v[i]]==ddx[x]+1){
vis[v[i]]=1;
if(my[v[i]]!=-1&&ddy[v[i]]==res)continue;
if(my[v[i]]==-1||dfs(my[v[i]])){
my[v[i]]=x;
mx[x]=v[i];
return true;
}
}
}
return false;
}
int maxMatch(){
int ans=0;
memset(mx,-1,sizeof(mx));
memset(my,-1,sizeof(my));
while(bfs()){
memset(vis,0,sizeof(vis));
for(int i=1;i<=n1;++i){
if(mx[i]==-1&&dfs(i))ans++;
}
}
return ans;
}
}Two;
void getprime(){
pcnt = 0;
memset(noprime, 0, sizeof(noprime));
noprime[0] = noprime[1] = 1;
for(int i = 2; i < N; ++i) {
if(!noprime[i])p[pcnt++] = i;
for(int j = 0; j < pcnt && i * p[j] < N; ++j) {
noprime[i * p[j]] = 1;
if(i % p[j] == 0)break;
}
}
}
void cal(int t,int id){
int tmp = (int)sqrt(t*1.0);
for(int i = 0; i < pcnt && p[i] <= tmp; ++i) {
if(t % p[i] == 0) {
have[p[i]].push_back(id);
while (t % p[i] == 0) {
t /= p[i];
}
}
if(t == 1)break;
}
if(t > 1)have[t].push_back(id);
}
int main(){
int tim;
getprime();
scanf("%d",&tim);
while(tim--){
scanf("%d", &n);
for(int i=0;i<N-2;++i)have[i].clear();
for(int i=1;i<=n;++i){
scanf("%d",&ar[i]);
cal(ar[i],i);
}
char s[2];
for(int i=1;i<=n;++i){
scanf("%s",s);
is[i]=0;
if(s[0]=='M')is[i]=1;
}
Two.clear();
Two.n1 = n; Two.n2 = n;
for(int i=0;i<N-2;++i){
if(have[i].size()<=1){
have[i].clear();
continue;
}
int len = have[i].size();
for(int j=0;j<len;++j){
for(int k=0;k<len;++k){
if(j==k)continue;
if(is[have[i][j]]!=is[have[i][k]]){
Two.add(have[i][j],have[i][k]);
}
}
}
have[i].clear();
}
printf("%d\n", n-Two.maxMatch()/2);
}
return 0;
}C题Intersections:
逆序对
AC代码
#include<bits/stdc++.h>
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long LL;
const int N = 1e5+7;
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
int n,q,tot;
int ar[N],bit[N],br[N];
void add(int x){
while(x<=n){
bit[x]++;
x += lowbit(x);
}
}
int query(int x){
int sum=0;
while(x){
sum+=bit[x];
x-=lowbit(x);
}
return sum;
}
int x,y;
int main(int argc, char const *argv[]){
int tim;
scanf("%d", &tim);
while(tim--){
scanf("%d", &n);
for(int i=1;i<=n;++i){
int x;scanf("%d",&x);
ar[x]=i;
}
for(int i=1;i<=n;++i){
int x;scanf("%d",&x);
br[i]=ar[x];
}
LL all=0;
memset(bit,0,sizeof(bit));
for(int i=1;i<=n;++i){
int p = br[i];
int tmp = query(p);
all+=tmp;
add(p);
}
all=n*1LL*(n-1)/2-all;
printf("%lld\n", all);
}
return 0;
}
E题Maximum Sum:
简单状压
AC代码
#include<bits/stdc++.h>
#define mme(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const int N = 10 + 8;
const int ME = 1<<16;
const int INF = 0x3f3f3f3f;
int n, m;
int ok[ME+1];
int dp[N][ME+1];
int ar[N][N];
inline bool adj(int x){
return (x&(x<<1))||(x&(x>>1));
}
inline int get_num(int t,int s){
int sum=0;
for(int i=0;i<n;++i){
if(s&(1<<i))sum+=ar[t][i];
}
return sum;
}
int main(){
int tim;
scanf("%d",&tim);
while(tim--){
scanf("%d", &n);
memset(dp, 0, sizeof(dp));
memset(ok,0,sizeof(ok));
int aa = 1<<n, tot = 0;
///单行合法状态储存进数组
for(int i = 0; i < aa; ++i){
if(adj(i) == 0)ok[tot++] = i;
}
for(int i=0;i<n;++i){
for(int j=0;j<n;++j){
scanf("%d",&ar[i][j]);
}
}
for(int _i = 0, i; _i < tot; ++_i){
i = ok[_i];
dp[0][_i]=get_num(0, i);
}
for(int _i = 0, i, j; _i < tot; ++_i){
i = ok[_i];
for(int _j = 0; _j < tot; ++_j){
j = ok[_j];
if((i&j)||(i&(j<<1))||(i&(j>>1)))continue;
dp[1][_i] = max(dp[0][_j]+get_num(1,i),dp[1][_i]);
}
}
for(int t = 2,i,j,k; t < n; ++t){
for(int _i = 0; _i < tot; ++_i){
i = ok[_i];
for(int _j = 0; _j < tot; ++_j){
j = ok[_j];
if((i&j)||(i&(j<<1))||(i&(j>>1)))continue;
dp[t][_i]=max(dp[t-1][_j]+get_num(t,i) ,dp[t][_i]);
}
}
}
int ans = 0;
for(int _i = 0, i, j; _i < tot; ++_i){
i = ok[_i];
ans=max(ans,dp[n-1][_i]);
}
printf("%d\n", ans);
}
return 0;
}G题Hard Equation:
扩展BSGS裸题：传送门
AC代码
#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
inline LL gcd(LL a, LL b) {
return (!b) ? a : gcd(b, a % b);
}
inline void Exgcd(LL a, LL b, LL &d, LL &x, LL &y) {
if (!b) { d = a, x = 1, y = 0; }
else { Exgcd(b, a % b, d, y, x), y -= x * (a / b); }
}
inline LL Solve(LL a, LL b, LL c) {// ax%c=b S.T. (a,c)=1
LL d, x, y;
Exgcd(a, c, d, x, y);
x = (x + c) % c;
return x * b % c;
}
inline LL Ksm(LL x, LL y, LL p) {
LL res = 1, t = x;
for(; y; y >>= 1) {
if (y & 1) res = res * t % p;
t = t * t % p;
}
return res;
}
#define mod 1313131
struct Hashset {
int head[mod], next[35010], f[35010], v[35010], ind;
void reset() {
ind = 0;
memset(head, -1, sizeof head);
}
void Insert(int x, int _v) {
int ins = x % mod;
for(int j = head[ins]; j != -1; j = next[j])
if (f[j] == x) {
v[j] = min(v[j], _v);
return;
}
f[ind] = x, v[ind] = _v;
next[ind] = head[ins], head[ins] = ind++;
}
int operator [] (const int &x) const {
int ins = x % mod;
for(int j = head[ins]; j != -1; j = next[j])
if (f[j] == x)
return v[j];
return -1;
}
}S;
LL BSGS(LL C, LL A, LL B, LL p) {// A^x%p=B S.T.(A,p)=1
if (p <= 100) {
LL d = 1;
for(int i = 0; i < p; ++i) {
if (d == B)
return i;
d = d * A % p;
}
return -1;
}
else {
int m = (int)sqrt(p);
S.reset();
LL d = 1, Search;
for(int i = 0; i < m; ++i) {
S.Insert(d, i);
d = d * A % p;
}
for(int i = 0; i * m < p; ++i) {
d = Ksm(A, i * m, p) * C % p;
Search = S[Solve(d, B, p)];
if (Search != -1)
return i * m + Search;
}
return -1;
}
}
int main() {
LL x, z, k;
register LL i, j;
int tim;
scanf("%d",&tim);
while(tim--){
scanf("%lld%lld%lld", &x, &k, &z);
LL d = 1;
bool find = 0;
for(i = 0; i < 100; ++i) {
if (d == k) {
printf("%lld\n", i);
find = 1;
break;
}
d = d * x % z;
}
if (find)continue;
LL t, C = 1, num = 0;
bool failed = 0;
while((t = gcd(x, z)) != 1) {
if (k % t != 0) {
failed = 1;
break;
}
z /= t;
k /= t;
C = C * x / t % z;
++num;
}
LL res = BSGS(C, x, k, z);
printf("%lld\n", res + num);
}
return 0;
}
K题Citations:
队友写的模拟贴个代码
AC代码
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cstring>
#include <cstdio>
#include <string>
#include <cmath>
#include <set>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<char, int> PCI;
const LL INF = 0x3f3f3f3f;
const LL LLINF = 0x3f3f3f3f3f3f3f3f;
const int MAX_N = (int)2e5+17;
const int MAX_M = (int)2e5+17;
int N, M, S, T, K, tp;
int a[MAX_N], b[MAX_N], lt[MAX_N];
int vis[MAX_N];
char s[21][322];
int main() {
//freopen("../data.in", "r", stdin);
//ios_base::sync_with_stdio(false);cin.tie(0);
//cin » T;
scanf("%d", &T);
int kase = 1;
while (T--) {
scanf("%d", &N);getchar();
while(N--) {
for(int i = 0; i < 10; i++) {
fgets(s[i], sizeof(s[i]), stdin);
//printf("%s",s[i]);
}
for(int i = 1; i <= 8; i++) {
if(s[i][0]=='a') {
vector<char> v;v.clear();
for(int j = 8; ; ) {
v.push_back(s[i][j]);
v.push_back(s[i][j+1]);
while(s[i][j] != ' ') j++;j++;
v.push_back(s[i][j]);
while(s[i][j] != ',' && s[i][j] != '}') j++;
if(s[i][j] == '}') break;
else j += 2;
}
for(int j = 0; j < v.size(); j+=3) {
printf("%c%c. %c%c ",v[j],v[j+1],v[j+2],(j+3<v.size())?',':'.');
}
break;
}
}
for(int i = 1; i <= 8; i++) {
if(s[i][0]=='t') {
for(int j = 7; s[i][j] != '}'; j++) {
printf("%c",s[i][j]);
}
printf(". ");
break;
}
}
for(int i = 1; i <= 8; i++) {
if(s[i][0]=='j') {
for(int j = 9; s[i][j] != '}'; j++) {
printf("%c",s[i][j]);
}
printf(". ");
break;
}
}
for(int i = 1; i <= 8; i++) {
if(s[i][0]=='y') {
for(int j = 6; s[i][j] != '}'; j++) {
printf("%c",s[i][j]);
}
printf(";");
break;
}
}
for(int i = 1; i <= 8; i++) {
if(s[i][0]=='v') {
for(int j = 8; s[i][j] != '}'; j++) {
printf("%c",s[i][j]);
}
printf("(");
break;
}
}
for(int i = 1; i <= 8; i++) {
if(s[i][0]=='n') {
for(int j = 8; s[i][j] != '}'; j++) {
printf("%c",s[i][j]);
}
printf("):");
break;
}
}
for(int i = 1; i <= 8; i++) {
if(s[i][0]=='p' && s[i][1]=='a') {
for(int j = 7; s[i][j] != '}'; j++) {
printf("%c",s[i][j]);
}
printf(".\n");
break;
}
}
}
}
return 0;
}