《省赛补题》

B：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<double,int> pii;
const int N = 1e6+5;
const int M = 1e6+5;
const LL Mod = 1e9+7;
#define rg register
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline int read()
    {
        int x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;

int a[N];
struct Node{int L,r,maxx;}node[N << 2];
void Pushup(int idx){node[idx].maxx = max(node[idx << 1].maxx,node[idx << 1 | 1].maxx);}
void build(int L,int r,int idx)
{
    node[idx].L = L,node[idx].r = r,node[idx].maxx = 0;
    if(L == r) return ;
    int mid = (L + r) >> 1;
    build(L,mid,idx << 1);
    build(mid + 1,r,idx << 1 | 1);
}
int query(int x,int idx)
{
    if(node[idx].L == node[idx].r) return node[idx].L;
    int mid = (node[idx].L + node[idx].r) >> 1;
    if(node[idx << 1].maxx >= x) return query(x,idx << 1);
    else return query(x,idx << 1 | 1);
}
void update(int x,int val,int idx)
{
    if(node[idx].L == node[idx].r)
    {
        node[idx].maxx += val;
        return ;
    }
    int mid = (node[idx].L + node[idx].r) >> 1;
    if(mid >= x) update(x,val,idx << 1);
    else update(x,val,idx << 1 | 1);
    Pushup(idx);
}
set<int> S;
map<int,int> mp;
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        int n,c;n = read(),c = read();
        for(rg int i = 1;i <= n;++i) a[i] = read();
        int num1 = 0,num2 = 0;
        build(1,n,1);
        for(rg int i = 1;i <= n;++i)
        {
            int pos = -1;
            if(node[1].maxx >= a[i]) pos = query(a[i],1);
            if(pos == -1) ++num1,update(num1,c - a[i],1);
            else update(pos,-a[i],1);
        }
        S.clear();mp.clear();
        for(rg int i = 1;i <= n;++i)
        {
            auto ip = S.end();
            if(S.size() == 0 || *(--ip) < a[i]) num2++,S.insert(c - a[i]),mp[c - a[i]]++;
            else
            {
                auto pos = S.lower_bound(a[i]);
                int tmp = *pos - a[i];
                mp[*pos]--;
                mp[tmp]++;
                if(mp[*pos] == 0) S.erase(*pos);
                if(mp[tmp] == 1) S.insert(tmp);
            }
        }
        printf("%d %d\n",num1,num2);
    }    
    system("pause");
    return 0;
}

C：

细节有点多。

注意的是这里的字典是M组（带着各自的代价）。

然后再去查询矩阵里的字符串。并且这个字符串必定满足连贯的非子串的完整字符串（且必须只能左到右读，或者上到下来读）

那么就可以构造字典树来查询。因为这里保证了总字典长度<= 4e6

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<double,int> pii;
const int N = 4e6+5;
const int M = 1e6+5;
const LL Mod = 1e9+7;
#define rg register
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline int read()
    {
        int x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;

string a[1005];
int tree[N][30],sum[N],k = 0;
void Insert(string s,int w)
{
    int len = s.size(),p = 0;
    for(rg int i = 0;i < len;++i)
    {
        int c = s[i] - 'a';
        if(!tree[p][c]) tree[p][c] = ++k;
        p = tree[p][c];
    }
    sum[p] = w;
}
int query(string s)
{
    int len = s.size(),p = 0;
    for(rg int i = 0;i < len;++i)
    {
        int c = s[i] - 'a';
        if(!tree[p][c]) return 0;
        p = tree[p][c];
    }
    return sum[p];
}
int main()
{
    ios :: sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    int ca;cin >> ca;
    while(ca--)
    {
        for(rg int i = 0;i <= k;++i) 
        {
            sum[i] = 0;
            for(rg int j = 0;j < 30;++j) tree[i][j] = 0;
        }
        k = 0;
        int n,m,f = 0;cin >> n >> m;
        for(rg int i = 0;i < n;++i) cin >> a[i];
        LL ans = 0;
        while(m--)
        {
            string s;
            int x;
            cin >> s >> x;
            Insert(s,x);
        }
        for(rg int i = 0;i < n;++i)//向右扫
        {
            if(f) break;
            int L = -1;
            string s = "";
            for(rg int j = 0;j < n;++j)
            {
                if(L == -1)
                {
                    if(a[i][j] != '#' && (j == 0 || a[i][j - 1] == '#'))
                    {
                        L = j,s += a[i][j];
                        if(j == n - 1) 
                        {
                            int x = query(s);
                            if(x == 0) f = 1;
                            else ans += x;
                            L = -1,s = "";
                        }
                    }
                }
                else
                {
                    if(a[i][j] == '#')
                    {
                        int x = query(s);
                        if(x == 0) f = 1;
                        else ans += x;
                        L = -1,s = "";
                    }
                    else if(j == n - 1)
                    {
                        s += a[i][j];
                        int x = query(s);
                        if(x == 0) f = 1;
                        else ans += x;
                        L = -1,s = "";
                    }
                    else s += a[i][j];
                }
            }
        }
        for(rg int j = 0;j < n;++j)//向下扫
        {
            if(f) break;
            int L = -1;
            string s = "";
            for(rg int i = 0;i < n;++i)
            {
                if(L == -1)
                {
                    if(a[i][j] != '#' && (i == 0 || a[i - 1][j] == '#')) 
                    {
                        L = i,s += a[i][j];
                        if(i == n - 1)
                        {
                            int x = query(s);
                            if(x == 0) f = 1;
                            else ans += x;
                            L = -1,s = "";
                        }
                    }
                }
                else
                {
                    if(a[i][j] == '#') 
                    {
                        int x = query(s);
                        if(x == 0) f = 1;
                        else ans += x;
                        L = -1,s = "";
                    }
                    else if(i == n - 1) 
                    {
                        s += a[i][j];
                        int x = query(s);
                        if(x == 0) f = 1;
                        else ans += x;
                        L = -1,s = "";
                    }
                    else s += a[i][j];
                }
            }
        }
        if(f) ans = -1;
        cout << ans << endl;
    }
    system("pause");
    return 0;
}

 E:

仔细读懂题的话，就是i^2的前缀和 + 区间的前k大的和。

直接上主席树就行。

i * i写成了a[i] * a[i]调了N久（笨比一个）

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<double,int> pii;
const int N = 1e5+5;
const int M = 1e6+5;
const LL Mod = 1e9+7;
#define rg register
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline int read()
    {
        int x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;

int rt[N],top = 0;
LL sum[N],a[N];
struct Node{int L,r;LL cnt,sum;}node[M * 20];
void Pushup(int idx)
{
    int ls = node[idx].L,rs = node[idx].r;
    node[idx].sum = node[ls].sum + node[rs].sum;
    node[idx].cnt = node[ls].cnt + node[rs].cnt;
}
int build(int L,int r)
{
    int idx = ++top;
    node[idx].cnt = node[idx].sum = 0;
    if(L == r) return idx;
    int mid = (L + r) >> 1;
    node[idx].L = build(L,mid);
    node[idx].r = build(mid + 1,r);
    return idx;
}
int Insert(int L,int r,int idx1,int x)
{
    int idx2 = ++top;
    node[idx2] = node[idx1];
    if(L == r)
    {
        node[idx2].cnt++;
        node[idx2].sum = node[idx2].cnt * L;
        return idx2;
    }
    int mid = (L + r) >> 1;
    if(mid >= x) node[idx2].L = Insert(L,mid,node[idx1].L,x);
    else node[idx2].r = Insert(mid + 1,r,node[idx1].r,x);
    Pushup(idx2);
    return idx2;
}
LL query(int x,int y,int L,int r,int k)
{
    if((node[y].cnt - node[x].cnt) <= k) return node[y].sum - node[x].sum;
    if(L == r) return 1LL * L * k;
    int mid = (L + r) >> 1;
    int ls1 = node[x].L,rs1 = node[x].r;
    int ls2 = node[y].L,rs2 = node[y].r;
    LL rcnt = node[rs2].cnt - node[rs1].cnt;
    if(rcnt >= k) return query(rs1,rs2,mid + 1,r,k);
    else return node[rs2].sum - node[rs1].sum + query(ls1,ls2,L,mid,k - rcnt);
}
int main()
{
    int ca;ca = read();
    while(ca--)
    {
        top = 0;
        int n;n = read();
        int mx = -1;
        for(rg int i = 1;i <= n;++i) a[i] = read(),mx = max(mx,(int)a[i]);
        for(rg int i = 1;i <= n;++i) sum[i] = sum[i - 1] + 1LL * i * i;
        rt[0] = build(1,mx);
        for(rg int i = 1;i <= n;++i) rt[i] = Insert(1,mx,rt[i - 1],a[i]);
        int q;q = read();
        while(q--)
        {
            int L,r,k;L = read(),r = read(),k = read();
            int m = r - L + 1;
            LL ans = sum[m];
            ans += query(rt[L - 1],rt[r],1,mx,k);
            printf("%lld\n",ans);
        }
    }
    system("pause");
    return 0;
}