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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#472913#8565. Basic Bloomsucup-team2307RE 1ms3892kbC++203.7kb2024-07-11 20:17:342024-07-11 20:17:39

Judging History

你现在查看的是最新测评结果

  • [2024-07-11 20:17:39]
  • 评测
  • 测评结果:RE
  • 用时:1ms
  • 内存:3892kb
  • [2024-07-11 20:17:34]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
//#define int ll
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
using pii = pair<int, int>;
using vi = vector<int>;
#define fi first
#define se second
#define pb push_back

const int MOD=998244353;

struct Big: vector<signed> {
    static const int BASE = MOD;
#define me (*this)
#define shrink(x) while (!x.empty() && !x.back()) x.pop_back();
    Big() {}
    Big(auto val) { while (val) pb(val % BASE), val /= BASE; }
    Big(const string& s) {
        for (int i = sz(s); i > 0; i -= 9)
            pb(stoi(i < 9 ? s.substr(0, i) : s.substr(i - 9, 9)));
        shrink(me);
    }
    friend ostream& operator<<(ostream& out, const Big& v) {
        out << (v.empty() ? 0 : v.back());
        for (int i = sz(v) - 2; i >= 0; i--)
            out << setw(9) << setfill('0') << v[i];
        return out;
    }
    auto operator<=>(const Big& oth) const {
        if (auto res = sz(me) <=> sz(oth); res != 0) return res;
        for (int i = sz(me) - 1; i >= 0; i--)
            if (auto res = me[i] <=> oth[i]; res != 0) return res;
        return strong_ordering::equal;
    }
#define pm(op,inv,cmp,end) \
        Big& operator op(const auto& oth) { \
            for (int i = 0, r = 0; i < sz(oth) || r; i++) { \
                if (i == size()) end(0); \
                me[i] op r + (i < sz(oth) ? oth[i] : 0); \
                if (r = me[i] cmp) me[i] inv BASE; \
            } \
            shrink(me); return me; \
        }
    pm(+=, -=, >= BASE, pb)
    pm(-=, +=, < 0, assert)
    Big operator*(const auto& oth) const {
        Big c; c.resize(sz(me) + sz(oth));
        rep(i, 0, sz(me))
            for (int j = 0, r = 0; j < sz(oth) || r; ++j) {
                ll cur = c[i + j] + me[i] * ll(j < sz(oth) ? oth[j] : 0) + r;
                c[i + j] = cur % BASE;
                r = cur / BASE;
            }
        shrink(c); return c;
    }
    int divmod(int oth) {
        int r = 0;
        for (int i = sz(me) - 1; i >= 0; i--) {
            ll cur = me[i] + (ll) r * BASE;
            me[i] = cur / oth;
            r = cur % oth;
        }
        shrink(me); return r;
    }
    Big& operator/=(int oth) {
        divmod(oth);
        return me;
    }
#define def(op, expr) friend auto operator op(Big a, const auto&b) { return a expr(b); }
    def(+, +=) def(-, -=) def(/, /=) def(%, .divmod)
    Big& operator*=(const auto& oth) { return me = me * oth; }
};

struct Small {
    Big::value_type val;
    Small(Big::value_type val) : val(val) { assert(val < Big::BASE); }
    static int size() { return 1; }
    auto operator[](int) const { return val; }
};

struct Seq
{
    int base,dig;
    Big num;
    Seq(int base,int dig):base(base),dig(dig),num{dig} {}
    void next()
    {
        num*=Big(base);
        num+=Big(dig);
    }
    friend bool operator<(const Seq&x,const Seq&y)
    {
        return x.num>y.num;
    }
};

const int N=2e3+100;

signed main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);

    priority_queue<Seq> pq;
    for(int base=2;base<=16;base++)
        for(int dig=1;dig<base;dig++)
            pq.emplace(base,dig);

    vector<int> nums{0};
    Big prv;
    while(nums.size()<=N)
    {
        Seq cur=pq.top();
        pq.pop();
        if(cur.num!=prv)
            prv=cur.num,
            nums.pb({prv.front()});
        cur.next();
        pq.push(cur);
    }
    for(int i=1;i<sz(nums);i++)
        (nums[i]+=nums[i-1])%=Big::BASE;
    int t;
    cin>>t;
    while(t--)
    {
        int l,r;
        cin>>l>>r;
        cout<<(nums[r]-nums[l-1]+Big::BASE)%Big::BASE<<"\n";
    }
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3892kb

input:

3
1 2
1 10
15 2000

output:

3
55
736374621

result:

ok 3 number(s): "3 55 736374621"

Test #2:

score: -100
Runtime Error

input:

100000
26 99975
57 99944
28 99973
62 99939
71 99930
25 99976
53 99948
60 99941
73 99928
72 99929
30 99971
7 99994
3 99998
35 99966
73 99928
68 99933
83 99918
37 99964
63 99938
17 99984
34 99967
74 99927
6 99995
3 99998
23 99978
91 99910
39 99962
85 99916
82 99919
17 99984
61 99940
31 99970
44 99957
...

output:


result: