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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
#445688#8527. Power Divisionsucup-team1231#WA 6ms14464kbC++202.5kb2024-06-16 08:20:102024-06-16 08:20:10

Judging History

你现在查看的是最新测评结果

  • [2024-06-16 08:20:10]
  • 评测
  • 测评结果:WA
  • 用时:6ms
  • 内存:14464kb
  • [2024-06-16 08:20:10]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
typedef long long int LLI;
typedef vector<int> vi;
typedef pair<int,int> pii;
#define MOD 1000000007
#define MOD2 998244353

int power[1001000],power2[1001000];
int a[300000];
vi vv[300001];
bitset<1001000> bs;
int getAll(int l,int r) {
    if (l == r) {
        vv[r+1].pb(l);
        return 0;
    }

    int i,mid = (l+r) / 2;
    getAll(l,mid),getAll(mid+1,r);
    vi all;
    int mx = 0;
    int sum1 = 0,sum2 = 0;
    map<pii,int> M1,M2;
    for (i = mid; i >= l; i--) {
        int u = a[i];
        while (bs[u]) bs[u] = 0,u++;
        bs[u] = 1,all.pb(u),mx = max(mx,u);
        sum1 += power[a[i]],sum1 %= MOD;
        sum2 += power2[a[i]],sum2 %= MOD2;
        int h1 = (power[mx+1]-sum1+MOD) % MOD;
        int h2 = (power2[mx+1]-sum2+MOD2) % MOD2;
        M1[mp(h1,h2)] = i;
    }
    sum1 = sum2 = 0;
    for (i = mid+1; i <= r; i++) {
        sum1 += power[a[i]],sum1 %= MOD;
        sum2 += power2[a[i]],sum2 %= MOD2;
        if (M1.count(mp(sum1,sum2))) vv[i+1].pb(M1[mp(sum1,sum2)]);
    }
    for (int u: all) bs[u] = 0;
    all.clear();
    mx = 0,sum1 = 0,sum2 = 0;
    for (i = mid+1; i <= r; i++) {
        int u = a[i];
        while (bs[u]) bs[u] = 0,u++;
        bs[u] = 1,all.pb(u),mx = max(mx,u);
        sum1 += power[a[i]],sum1 %= MOD;
        sum2 += power2[a[i]],sum2 %= MOD2;
        int h1 = (power[mx+1]-sum1+MOD) % MOD;
        int h2 = (power2[mx+1]-sum2+MOD2) % MOD2;
        M2[mp(h1,h2)] = i;
    }
    sum1 = sum2 = 0;
    for (i = mid; i >= l; i--) {
        sum1 += power[a[i]],sum1 %= MOD;
        sum2 += power2[a[i]],sum2 %= MOD2;
        if (M2.count(mp(sum1,sum2))) vv[M2[mp(sum1,sum2)]+1].pb(i);
    }
    for (int u: all) bs[u] = 0;
    all.clear();
    return 0;
}
int dp[300001];
int main() {
    int i;
    power[0] = power2[0] = 1;
    for (i = 1; i < 1001000; i++) {
        power[i] = (2*power[i-1]) % MOD;
        power2[i] = (2*power2[i-1]) % MOD2;
    }
    int n;
    scanf("%d",&n);
    for (i = 0; i < n; i++) scanf("%d",&a[i]);

    dp[0] = 1;
    getAll(0,n-1);
    for (i = 0; i < n; i++) {
        sort(vv[i+1].begin(),vv[i+1].end());
        vv[i+1].resize(unique(vv[i+1].begin(),vv[i+1].end())-vv[i+1].begin());
        for (int u: vv[i+1]) {
            dp[i+1] += dp[u];
            if (dp[i+1] >= MOD) dp[i+1] -= MOD;
        }
    }
    printf("%d\n",dp[n]);

    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 6ms
memory: 13336kb

input:

5
2 0 0 1 1

output:

6

result:

ok 1 number(s): "6"

Test #2:

score: 0
Accepted
time: 5ms
memory: 14464kb

input:

1
0

output:

1

result:

ok 1 number(s): "1"

Test #3:

score: 0
Accepted
time: 2ms
memory: 13640kb

input:

2
1 1

output:

2

result:

ok 1 number(s): "2"

Test #4:

score: 0
Accepted
time: 0ms
memory: 13692kb

input:

3
2 1 1

output:

3

result:

ok 1 number(s): "3"

Test #5:

score: -100
Wrong Answer
time: 2ms
memory: 13972kb

input:

4
3 2 2 3

output:

3

result:

wrong answer 1st numbers differ - expected: '4', found: '3'