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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
#758322#9619. 乘积,欧拉函数,求和electricstick#WA 312ms2692kbC++202.1kb2024-11-17 17:44:302024-11-17 17:44:32

Judging History

你现在查看的是最新测评结果

  • [2024-11-17 17:44:32]
  • 评测
  • 测评结果:WA
  • 用时:312ms
  • 内存:2692kb
  • [2024-11-17 17:44:30]
  • 提交

answer

#include<cstdio>
#include<algorithm>
const int N = 3010, mod = 998244353, Sq = 55, M = 16;
int n, a[N];
int p[M] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};
int f[2][(1 << M) + 10][2];
int inv[N];
struct num {
    int stu;
    int bigp;
    int data;
    bool operator< (const num &oth) {
        return bigp < oth.bigp;
    }
}c[N];
int main() {
    inv[1] = 1;
    for(int i = 2; i <= 3000; i++) {
        inv[i] = mod - 1ll * inv[mod % i] * (mod / i) % mod;
    }
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        c[i].data = a[i];
        int x = a[i], y = 0;
        for(int j = 0; j < 16; j++) {
            if(x % p[j] == 0) {
                y |= (1 << j);
                while(x % p[j] == 0) x /= p[j];
            }
        }
        c[i].bigp = x;
        c[i].stu = y;
    }
    std::sort(c + 1, c + 1 + n);
    f[0][0][0] = 1;
    for(int i = 1; i <= n; i++) {
        if(c[i].bigp != c[i - 1].bigp) {
            for(int j = 0; j < (1 << 16); j++) {
                f[i&1^1][j][0] = (f[i&1^1][j][0]+1ll*f[i&1^1][j][1]*(c[i-1].bigp==1?1:inv[c[i-1].bigp]%mod*(c[i-1].bigp-1)))%mod;
                f[i&1^1][j][1] = 0;
            }
        }
        for(int j = 0; j < (1 << 16); j++) {
            f[i&1][j][0] = f[i&1^1][j][0];
            f[i&1][j][1] = f[i&1^1][j][1];
        }
        for(int j = 0; j < (1 << 16); j++) {
            f[i&1][j|c[i].stu][1] = (f[i&1][j|c[i].stu][1] + 1ll*(f[i&1^1][j][0]+f[i&1^1][j][1])*c[i].data)%mod;
        }
        // printf("##%d: %d %d\n", i, c[i].stu, c[i].data);
        // for(int j = 0; j < (1 << 3); j++) {
        //     printf("%d %d\n", j, f[i&1][j][1]);
        // }
    }
    int ans = 0;
    for(int j = 0; j < (1 << 16); j++) {
        int now = (f[n&1][j][0] + 1ll*f[n&1][j][1]*(c[n].bigp==1?1:inv[c[n].bigp]%mod*(c[n].bigp-1)))%mod;
        // printf("%d %d\n", j, now);
        for(int k = 0; k < 16; k++) {
            if((j >> k) & 1) {
                now = 1ll * now * inv[p[k]] % mod * (p[k] - 1) % mod;
            }
        }
        ans = (ans + now) % mod;
    }
    printf("%d\n", ans);
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 6ms
memory: 2636kb

input:

5
1 6 8 6 2

output:

892

result:

ok single line: '892'

Test #2:

score: 0
Accepted
time: 6ms
memory: 2692kb

input:

5
3 8 3 7 8

output:

3157

result:

ok single line: '3157'

Test #3:

score: -100
Wrong Answer
time: 312ms
memory: 2600kb

input:

2000
79 1 1 1 1 1 1 2803 1 1 1 1 1 1 1609 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2137 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 613 1 499 1 211 1 2927 1 1 1327 1 1 1123 1 907 1 2543 1 1 1 311 2683 1 1 1 1 2963 1 1 1 641 761 1 1 1 1 1 1 1 1 1 1 1 1489 2857 1 1 1 1 1 1 1 1 1 1 1 1 1 967 1 821 1 1 1 1 2143 1861...

output:

-30518812

result:

wrong answer 1st lines differ - expected: '50965652', found: '-30518812'