2020 China Collegiate Programming Contest Mianyang Site

2020 China Collegiate Programming Contest Mianyang Site

#include <bits/stdc++.h>

#define int ll
//#pragma GCC optimize(2)
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
const int mod=998244353;

int a[maxn];
int t=0;
int n;

//check(x) 考虑当前得x是否可以对所有小于x得所有数都添加到x即可，
// 由于操作x次，所以当a[i]过小时，可能出现check时负溢出。
bool check(int x){
    
    
    int res=x;
    for(int i=1;i<=n;i++){
    
    
        if(a[i]>=x) continue;
        res-=x-a[i];
        if(res<=0) break;
    }
    return res>0;
}

void solve() {
    
    
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i],++a[i];
    int l=0,r=1e18;
    //二分答案。
    while (l<r){
    
    
        int mid = (r + l + 1) >> 1;
        if(check(mid)) l=mid;
        else r=mid-1;
    }
    cout<<"Case #"<<t<<": "<<l<<"\n";
}

signed main() {
    
    
    //ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
    cin >> _;
    while (_--) {
    
    
        t++;
        solve();
    }
    return 0;
}

7-11 Knowledge is Power

#include <bits/stdc++.h>

#define int ll
//#pragma GCC optimize(2)
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
const int mod=998244353;

int t=0;

int gcd(int a,int b){
    
    
    return b==0?a:gcd(b,a%b);
}

void solve() {
    
    
    int x,res;
    cin>>x;
    if(x&1) res=1;
    else if(x==6) res=-1;
    else{
    
    
        int tx=x/2;
        if(!(tx&1)) res=2;
        else{
    
    
            res=4;
            if(x%3==0) res=min(res,2ll);
            int tmp=x-4;
            if(tmp%3==0&&(tmp/3)%2==0&&tmp/3>1){
    
    
                int a=tmp/3,b=a+1,c=a+3;
                if(gcd(a, b) == 1 && gcd(a, c) == 1 && gcd(b, c) == 1) res = min(res, 3ll);
            }
            --tmp;
            if(tmp%3==0&&(tmp/3)%2&&tmp/3>1){
    
    
                int a=tmp/3,b=a+2,c=a+3;
                if(gcd(a, b) == 1 && gcd(a, c) == 1 && gcd(b, c) == 1) res = min(res, 3ll);
            }
        }
    }
    printf("Case #%d: %d\n", t, res);
}

signed main() {
    
    
    //ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
    cin >> _;
    while (_--) {
    
    
        t++;
        solve();
    }
    return 0;
}

7-10 Joy of Handcraft

#include <bits/stdc++.h>

#define int ll
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int maxn = 1e5 + 10;


int tree[maxn * 4];
int lazy[maxn * 4];
int ca = 0;

struct node {
    
    
    int t, x;
} a[maxn], b[maxn];

void init() {
    
    
    ca++;
    memset(tree, 0, sizeof(tree));
    memset(lazy, 0, sizeof(lazy));
}

bool cmp(node a, node b) {
    
    
    if (a.t == b.t) return a.x > b.x;
    return a.t < b.t;
}

void pushdown(int t) {
    
    
    lazy[t << 1] = max(lazy[t], lazy[t << 1]);
    lazy[t << 1 | 1] = max(lazy[t], lazy[t << 1 | 1]);
    tree[t << 1] = max(lazy[t << 1], tree[t << 1]);
    tree[t << 1|1] = max(lazy[t << 1 | 1], tree[t << 1 | 1]);
}

int query(int l, int r, int t, int L, int R) {
    
    
    if (l >= L && r <= R) return tree[t];
    if (l > R || r < L) return 0;
    int mid = (l + r) >> 1;
    int d = 0;
    pushdown(t);
    if (mid >= L) d = max(d, query(l, mid, t << 1, L, R));
    if (mid < R) d = max(d, query(mid + 1, r, t << 1 | 1, L, R));
    return d;
}

void update(int l, int r, int t, int L, int R, int d) {
    
    
    if (l >= L && r <= R) lazy[t] = max(lazy[t], d), tree[t] = max(tree[t], lazy[t]);
    else {
    
    
        pushdown(t);
        int mid = (l + r) >> 1;
        if (mid >= L)update(l, mid, t << 1, L, R, d);
        if (mid < R) update(mid + 1, r, t << 1 | 1, L, R, d);
        //tree[t]=max(tree[t<<1],tree[t<<1|1]);
    }
}



void solve() {
    
    
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) cin >> b[i].t >> b[i].x;
    sort(b + 1, b + n + 1, cmp);
    int top = 1;
    a[1] = b[1];
    for (int i = 2; i <= n; ++i)
        if (b[i].t != a[top].t)
            a[++top] = b[i];
    for (int i = 1; i <= top; ++i) {
    
    
        for (int k = 0;; ++k) {
    
    
            int l = 2 * k * a[i].t + 1;
            int r = min(l - 1 + a[i].t, m);
            if (l <= m && r <= m) update(1, m, 1, l, r, a[i].x);
            if (l >= m || r >= m) break;
        }
    };
    printf("Case #%d:", ca);
    for (int i = 1; i <= m; ++i)
        cout << " " << query(1, m, 1, i, i);
    cout << "\n";
}

signed main() {
    
    
    //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int _ = 1;
    cin >> _;
    while (_--) {
    
    
        init();
        solve();
    }
    return 0;
}