Diary_2.28

codeforces

3. Limited Repainting

const int maxn = 3e5+10;
int n, k, a[maxn];
string s;
bool check(int mx){
    int len = 0, cnt = 0;
    bool not_need = true;
    for(int i=0; i<n; i++){
        if(a[i] > mx){
            if(s[i] == 'R'){
                if(len && not_need == false)  cnt++;
                not_need = true;
                len = 0;
            }else{
                not_need = false;
                len++;
            }
        }else  len++;
    }
    if(len && not_need == false)    cnt++;
    if(cnt > k) return false;
    else    return true;
}
void solve(){
    cin >> n >> k >> s;
    int mx = 0;
    for(int i=0; i<n; i++){
        cin >> a[i];
        mx = max(mx, a[i]);
    }
    int l = 0, r = mx, ans = -1;
    while(l <= r){
        int mid = (l+r)/2;
        if(check(mid)){
            r = mid-1;
            ans = mid;
        }else  l = mid+1;
    }
    if(ans == -1)   cout << "ERROR" << endl;
    else    cout << ans << endl;
}

4. 砍树

#define int long long
const int maxn = 1e6+10;
int n, m, a[maxn];
bool check(int mx){
    int ans = 0;
    for(int i=0; i<n; i++){
        if(a[i] > mx){
            ans += a[i]-mx;
        }
    }
    if(ans >= m) return true;
    else    return false;
}
void solve(){
    cin >> n >> m;
    int mx = 0;
    for(int i=0; i<n ;i++){
        cin >> a[i];
        mx = max(mx, a[i]);
    }
    int l = 0, r = mx, ans = -1;
    while(l <= r){
        int mid = (l+r)>>1;
        if(check(mid)){
            ans = mid;
            l = mid+1;
        }else{
            r = mid-1;
        }
    }
    if(ans == -1)   cout << "ERROR" << endl;
    else    cout << ans << endl;
}

5. Skibidus and Fanum Tax (hard version)

#define int long long
const int maxn = 2e5+10;
int l[maxn], r[maxn], n, m, inf = 1e17;
void solve(){
    cin >> n >> m;
    for(int i=1; i<=n; i++)  cin >> l[i];
    for(int k=1; k<=m; k++)  cin >> r[k];
    sort(r, r+m+1);
    l[0] = -inf;
    for(int i=1; i<=n; i++){
        int opl = l[i], opr;
        int pos = lower_bound(r+1, r+m+1, l[i]+l[i-1])-r;
        if(pos <= m && pos >= 0)    opr = r[pos]-l[i];
        else    opr = inf;
        if(opl >= l[i-1])   opl = l[i];
        else    opl = inf;
        l[i] = min(opl, opr);
    }
    if(l[n] >= inf){
        cout << "NO" << endl;
    }else{
        cout << "YES" << endl;
    }
}

6. Counting Pairs

#define int long long
const int maxn = 2e5+10;
int a[maxn], n, x, y;
void solve(){
    cin >> n >> x >> y;
    int sum = 0;
    for(int i=0; i<n; i++){
        cin >> a[i];
        sum += a[i];
    }
    sort(a, a+n);
    int ans = 0;
    // sum-(a[i]+a[j]) <= y   找右边界  
    // sum-a[i]-a[j] <= y
    // a[j] >= sum-a[i]-y
    // sum-(a[i]+a[j]) >= x   找左边界
    // sum-a[i]-a[j] >= x
    // a[j] <= sum-a[i]-x
    for(int i=0; i<n; i++){
        int pl = lower_bound(a, a+n, sum-y-a[i])-a;
        int pr = upper_bound(a, a+n, sum-a[i]-x)-a-1;
        // cout << pl << ' ' << pr << endl;
        if(pl!=n && pr!=n){
            if(pr != -1)
                ans += pr-pl+1;
            if(a[i] >= a[pl] && a[i] <= a[pr])  ans--;
        }
    }
    cout << ans/2 << endl;
}

突然发现而非几乎没练过，我要爆学二分！！！