Ozon Tech Challenge 2020 C. Kuroni and Impossible Calculation 模数
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题意:
给出一个数列
a
[
]
a[]
a[],以及一个模数
m
m
m,计算
∏
1
<
=
i
<
j
<
=
n
∣
a
[
i
]
−
a
[
j
]
∣
\\prod_1 <= i < j <= n \\left|a[i] - a[j]\\right|
∏1<=i<j<=n∣a[i]−a[j]∣.
数据范围:
1
<
=
n
<
=
2
∗
1
0
5
,
1
<
=
m
<
=
1000
,
0
<
=
a
[
i
]
<
=
1
0
9
1 <= n <= 2 * 10 ^ 5, 1 <= m <= 1000, 0 <= a[i] <= 10^9
1<=n<=2∗105,1<=m<=1000,0<=a[i]<=109.
思路:
很容易想到的是
O
(
n
2
)
O(n^2)
O(n2)的做法,但是看到
n
n
n的范围,明显会超时。
又会看到
m
m
m的范围,不是一个固定的非常大的质数当作模数,所以
m
m
m应该是突破口。
根据
m
m
m,可以分成两种情况。
- n < = m n <= m n<=m, 这样 n n n的范围小于 1000 1000 1000,可以用 O ( n 2 ) O(n ^ 2) O(n2)的算法解决。
-
n
>
m
n > m
n>m,模完
m
m
m之后最多有
m
m
m个不同的模数,而数列中的数的个数大于
m
m
m,一定至少存在两个数,
a
[
i
]
≡
a
[
j
]
m
o
d
m
a[i] \\equiv a[j] \\mod m
a[i]≡a[j]modm, 那么
∣
a
[
i
]
−
a
[
j
]
∣
≡
0
m
o
d
m
|a[i] - a[j]| \\equiv 0 \\mod m
∣a[i]−a[j]∣≡0modm.
最终的乘积就是0.
代码:
/**
* Author : Xiuchen
* Date : 2020-03-03-23.20.52
* Description : C.cpp
*/
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<cmath>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 2e5 + 100;
int gcd(int a, int b)
return b ? gcd(b, a % b) : a;
ll m, n;
ll a[maxn];
int main()
scanf("%lld%lld", &n, &m);
for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);
if(n <= m)
ll ans = 1 % m;
for(int i = 1; i <= n; i++)
for(int j = i + 1; j <= n; j++)
ans = (ans * abs(a[i] - a[j])) % m;
printf("%lld\\n", ans);
else printf("0\\n");
return 0;
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