# CodeForces release C Prefix Product Sequence, codeforces487c

Source: Internet
Author: User

CodeForces release C Prefix Product Sequence, codeforces487c

Question:

Construct a 1 ~ The arrangement of n makes n prefix product % n A 0 ~ N-1 Arrangement

Ideas:

First, confirm that n must be put. Otherwise, % n will have multiple zeros. Then, the prefix product at the n-1 position is (n-1 )!

Next we will discuss that when n is the Union number, only n = 4 has a solution, because if n is the Union number, it can be split into p * q in the form of obviously pq | (n-1 )!

Then construct ai = (I + 1) * inv [I] because (I + 1) * inv [I] = (j + 1) * inv [j] Must Have I = j. Therefore, this construction satisfies the need for ai to be unique, that is, an arrangement.

In addition, this structure makes the prefix a1 * a2 * a3... = 1*2 * inv [1] * 3 * inv [2]... then the result of % n is also an arrangement.

The final answer is 1 ~ The inverse element of n can be used as a table to find the recurrence formula (mod-mod/I) * inv [mod % I] % mod

PS: teammates said that the idea of construction comes from "this question % n must be very special, so try 1 2 3 4 5"

Code:

`#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<map>#include<set>#include<vector>#include<queue>#include<cstdlib>#include<ctime>#include<cmath>using namespace std;typedef long long LL;#define N 100010int inv[N];int n;int main() {    scanf("%d", &n);    if (n == 1) {        puts("YES");        printf("1\n");        return 0;    } else if (n == 4) {        puts("YES");        printf("1\n3\n2\n4\n");        return 0;    }    for (int i = 2; i * i <= n; i++) {        if (n % i == 0) {            puts("NO");            return 0;        }    }    puts("YES");    puts("1");    inv[1] = 1;    for (int i = 2; i < n; i++) {        inv[i] = (LL) (n - n / i) * inv[n % i] % n;        printf("%d\n", (int) ((LL) (i) * inv[i - 1] % n));    }    printf("%d\n", n);    return 0;}`

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