HDU6440 Dream(费马小定理+构造) -2018CCPC网络赛1003

题意：

给定素数p，定义p内封闭的加法和乘法，使得$(m+n)^p=m^p+n^p$

思路：

由费马小定理，p是素数，$a^{p-1}\equiv 1(mod\;p)$

所以$(m+n)^{p}\equiv (m+n)(mod\;p)$

$m^{p}\equiv m(mod\;p)$

$n^{p}\equiv n(mod\;p)$

所以在模意义下，有$(m+n)^p=m^p+n^p$

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e5+2;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int main() {
    int T;
    scanf("%d", &T);
    while(T--){
        ll n;
        scanf("%I64d", &n);
            for(ll i = 0; i < n; i++){
                for(ll j = 0; j < n; j++){
                    printf("%I64d ", (ll)(i+j)%n);
                }
                printf("\n");
            }
            for(ll i = 0; i < n; i++){
                for(ll j = 0; j < n; j++){

                    printf("%I64d ", (ll)i*j%n);
                }
                printf("\n");
            }

    }
    return 0;
}

原文地址：https://www.cnblogs.com/wrjlinkkkkkk/p/9538508.html