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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
#709726#6513. Expression 3MiniLongTL 1ms11808kbC++203.2kb2024-11-04 16:28:462024-11-04 16:28:47

Judging History

你现在查看的是最新测评结果

  • [2024-11-04 16:28:47]
  • 评测
  • 测评结果:TL
  • 用时:1ms
  • 内存:11808kb
  • [2024-11-04 16:28:46]
  • 提交

answer

#include <bits/stdc++.h>
#define _rep(i, x, y) for(int i = x; i <= y; ++i)
#define _req(i, x, y) for(int i = x; i >= y; --i)
#define _rev(i, u) for(int i = head[u]; i; i = e[i].nxt)
#define pb push_back
#define fi first
#define se second
#define mst(f, i) memset(f, i, sizeof f)
using namespace std;
#ifdef ONLINE_JUDGE
#define debug(...) 0
#else
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#endif
typedef long long ll;
typedef pair<int, int> PII;
namespace fastio{
    #ifdef ONLINE_JUDGE
    char ibuf[1 << 20],*p1 = ibuf, *p2 = ibuf;
    #define get() p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++
    #else
    #define get() getchar()
    #endif
    template<typename T> inline void read(T &t){
        T x = 0, f = 1;
        char c = getchar();
        while(!isdigit(c)){
            if(c == '-') f = -f;
            c = getchar();
        }
        while(isdigit(c)) x = x * 10 + c - '0', c = getchar();
        t = x * f;
    }
    template<typename T, typename ... Args> inline void read(T &t, Args&... args){
        read(t);
        read(args...);
    }
    template<typename T> void write(T t){
        if(t < 0) putchar('-'), t = -t;
        if(t >= 10) write(t / 10);
        putchar(t % 10 + '0');
    }
    template<typename T, typename ... Args> void write(T t, Args... args){
        write(t), putchar(' '), write(args...);
    }
    template<typename T> void writeln(T t){
        write(t);
        puts("");
    }
    template<typename T> void writes(T t){
        write(t), putchar(' ');
    }
    #undef get
};
using namespace fastio;
#define multitest() int T; read(T); _rep(tCase, 1, T)
namespace Calculation{
    const ll mod = 998244353;
    ll ksm(ll p, ll h){ll base = p % mod, res = 1; while(h){if(h & 1ll) res = res * base % mod; base = base * base % mod, h >>= 1ll;} return res;}
    void dec(ll &x, ll y){x = ((x - y) % mod + mod) % mod;}
    void add(ll &x, ll y){x = (x + y) % mod;}
    void mul(ll &x, ll y){x = x * y % mod;}
    ll sub(ll x, ll y){return ((x - y) % mod + mod) % mod;}
    ll pls(ll x, ll y){return ((x + y) % mod + mod) % mod;}
    ll mult(ll x, ll y){return x * y % mod;}
}
using namespace Calculation;
const int N = 2e5 + 5;
ll n, a[N], fac[N], ifac[N], b[N];
char s[N];
ll f[N][2];
ll C(ll n, ll m){
    if(n < m || m < 0) return 0;
    if(n == m || !m) return 1;
    return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
int main(){
    read(n);
    fac[0] = ifac[0] = fac[1] = ifac[1] = 1;
    _rep(i, 2, n) fac[i] = fac[i - 1] * i % mod, ifac[i] = (mod - mod / i) * ifac[mod % i] % mod;
    _rep(i, 2, n) ifac[i] = ifac[i - 1] * ifac[i] % mod;
    _rep(i, 1, n) read(a[i]); scanf("%s", s + 1);
    _rep(i, 1, n - 1) b[i] = s[i] == '+' ? 0 : 1;
    ll ans = 0;
    _rep(i, 1, n){
        f[i][0] = 1, f[i][1] = 0;
        _req(j, i - 1, 1){
            _rep(k, 0, 1) f[j][k] = (f[j + 1][k ^ b[j]] + f[j + 1][k] * (i - j - 1) % mod) % mod;
        }
        ll cur = C(n - 1, i - 1) * fac[n - i] % mod;
        add(ans, f[1][0] * cur % mod * a[i] % mod + f[1][1] * cur % mod * (mod - a[i]) % mod);
    }
    writeln(ans);
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 9716kb

input:

4
9 1 4 1
-+-

output:

46

result:

ok 1 number(s): "46"

Test #2:

score: 0
Accepted
time: 0ms
memory: 11808kb

input:

5
1 2 3 4 5
+-+-

output:

998244313

result:

ok 1 number(s): "998244313"

Test #3:

score: -100
Time Limit Exceeded

input:

100000
664815434 205025136 871445392 797947979 379688564 336946672 231295524 401655676 526374414 670533644 156882283 372427821 700299596 166140732 677498490 44858761 185182210 559696133 813911251 842364231 681916958 114039865 222372111 784286397 437994571 152137641 650875922 613727135 209302742 5321...

output:


result: