上海理工大学联想杯 4题滚粗记

果然是被生活压垮的人，即将退役和告别大学生活了

早上放下了GRE来看题，写了两个小时就又不得不去学gre了，希望330+

比赛开始20分钟之后才到自习室，然后这个时候排名已经1100开外了

第一题，一看，问最长的等差数列长度，给了公差，所以拿map记录一下就好了

#include <bits/stdc++.h>
using namespace std;
#define limit (2000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
int n, m,k;
int a[limit];
void solve(){
    cin>>n>>k;
    map<int, int>mp;
    rep(i,1,n){
        cin>>a[i];
    }
    sort(a + 1, a + 1 + n);
    ll ans = 1;
    rep(i,1,n){
        if(!mp.count(a[i])){
            mp[a[i]] = mp[a[i] - k] + 1;
            ans = max(ans, 1ll * mp[a[i]]);
        }
    }
    cout<<ans<<endl;

}
int32_t main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    //FASTIO

    //kase = read();
    //while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}

然后看了下K，问最大匹配，数据还小得一批，笑了，暴力加边，dinic水了过去

#include <bits/stdc++.h>
using namespace std;
#define limit (200000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
int n,m,vs,ve,p;
int layer[limit],head[limit], cnt;
struct node{
    int to ,next;
    ll flow, w;
}edge[limit];
ll max_flow;
void add_one(int u , int v, ll flow = 0){
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    edge[cnt].flow = flow;
    edge[cnt].w = 0;
    head[u] = cnt++;
}
inline void add(int u, int v, ll flow){
    add_one(u,v,flow);
    add_one(v, u,0);
}
inline void init(bool flag = true){
    if(flag){
        memset(head, -1, sizeof(head));
        cnt = 0;
    }else{
        memset(layer, -1, sizeof(layer));
    }
}
inline bool bfs(){
    init(false);
    queue<int>q;
    layer[vs] = 0;//从第0层开始
    q.push(vs);
    while (q.size()){
        int u = q.front();
        q.pop();
        traverse(u){
            int v = edge[i].to,flow = edge[i].flow;
            if(layer[v] == -1 && flow > 0){
                layer[v] = layer[u] + 1;//迭代加深
                q.push(v);
            }
        }
    }
    return ~layer[ve];
}
ll dfs(int u, ll flow){
    if(u == ve)return flow;
    ll rev_flow = 0,min_flow;
    traverse(u){
        int v =edge[i].to;
        ll t_flow = edge[i].flow;
        if(layer[v] == layer[u] + 1 && t_flow > 0){
            min_flow = dfs(v, min(flow, t_flow));
            flow -= min_flow;
            edge[i].flow -= min_flow;
            rev_flow += min_flow;
            edge[i^1].flow += min_flow;
            if(!flow)break;
        }
    }
    if(!rev_flow)layer[u] = -1;
    return rev_flow;
}
void dinic(){
    while (bfs()){
        max_flow += dfs(vs,inf);
    }
}
int a[limit], b[limit];
void solve(){
    init();
    n = read();
    vs = 90001, ve = vs + 1;
    rep(i,1,n){
        a[i] = read();
        add(vs, i, 1);
    }
    rep(i,1,n){
        b[i] = read();
        add(i + n, ve, 1);
    }
    rep(i,1,n){
        rep(j,1,n){
            if(__gcd(a[i] , b[j]) != 1){
                add(i, j + n, INF);
            }

        }
    }
    dinic();
    write(max_flow);

}
int32_t main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    //FASTIO

    //kase = read();
    //while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}

然后C题，问匹配，好像列出来之后只有6种情况能对答案产生贡献，那么就手动模拟一下就好了

#include <bits/stdc++.h>
using namespace std;
#define limit (2000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
int n,m,k;
int ca,cat,c,at,t,a;
void solve(){
    cin>>n;
    rep(i,1,n) {
        string str;
        cin >> str;
        if (str == "ca")++ca;
        if (str == "cat")++cat;
        if (str == "a")++a;
        if (str == "at")++at;
        if (str == "t")++t;
        if (str == "c")++c;
    }
    // ca + t
    ll ans = min(ca, t), tmp;
    t -= ans, ca -= ans;
    // c + at
    tmp = min(c, at); ans += tmp;
    c -= tmp, at -= tmp;
    // cat
    ans += cat, cat = 0;
    // c + a + t
    tmp = min({c , a , t}); ans += tmp;
    c -= tmp, a -= tmp, t -= tmp;
    cout<<ans<<endl;

}
int32_t main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    FASTIO

    //kase = read();
    //while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}

然后看了看B，发现题意太过复杂，虽然感觉不难，但是念及自己可怜的分数还是算了，开了J

然后发现J的范围是1e7，很典型的素数筛题，想到根号分解数字，找到最小质因数，然后log计算贡献，再从增量里把这些剪掉就行，暴力加答案。

然后好家伙，tle了，我跟队友说不要写了，这题有坑，然后队友过了一会儿告诉我，tle个锤子，他过了。

然后我发现，好像__int128没关上，那没事了，果断提交并且AC，话说我这个__int128为了谨慎起见还开了fread，没想到这么毒瘤

#include <bits/stdc++.h>
using namespace std;
#define limit (20000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
ll n,m, k;
ll prime[limit], num[limit],tot;
int a[limit], b[limit];
void get_prime( const int &N=2e7){
    memset(num, 1, sizeof(num));
    tot = 0;
    rep(i ,2,N){
        if(num[i])prime[++tot] = i;
        for(int j = 1 ; j <= tot && i * prime[j] <= N ; ++j){
            num[i * prime[j]] = 0;
            if(i % prime[j] == 0)break;//线性筛
        }
    }
}
void process(ll x){
    if(num[x]){
        a[x] = x;
        b[x] = 1;
        return;
    }
    for(ll i = 2 ; i * i <= x; ++i){
        if(num[i] and x % i == 0){
            a[x] = i;
            ll tmp = x;
            while(tmp % i == 0){
                b[x]++;
                tmp /= i;
            }
            return;
        }
    }


}
void solve(){
    get_prime();
    n = read(),m = read(), k= read();
    rep(i,1 + k,n + k){
        process(i);
    }
    ll ans = 0;
    rep(i,1 + k,n + k){
        ll ti = b[i] - (b[i] / m);
        ll tmp = i;
        rep(j,1,ti){
            tmp /= a[i];
        }
        ans += i - tmp;
    }
    write(ans);
}
int32_t main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    FASTIO

    //kase = read();
    //while (kase--)
    solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}

然后看了看I题，好像三分写假了，那就撤退好了，今天还有新的任务呢。

感觉之后还是多一些训练，希望有生之年能够成全我一次吧，今年ICPC加油！