Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1014 Accepted Submission(s): 200
Special Judge
Problem Description
Freshmen frequently make an error in computing the power of a sum of real numbers, which usually origins from an incorrect equation p is a prime, the identity
holds true for every pair of non-negative integers p. Power is defined as
Obviously there exists an extremely simple solution that makes all operation just produce zero. So an extra constraint should be satisfied that there exists an integer p ought to be closed under the operation of your definitions.
Hint for sample input and output:
From the table we get 1.
They are the same.
p
holds true for every pair of non-negative integers p. Power is defined as
0
Obviously there exists an extremely simple solution that makes all operation just produce zero. So an extra constraint should be satisfied that there exists an integer p ought to be closed under the operation of your definitions.
Hint
Hint for sample input and output:
From the table we get 1.
They are the same.
Input
The first line of the input contains an positive integer p is guranteed to be a prime.
Output
For each test case, you should print ).
Sample Input
1
2
Sample Output
0 1
1 0
0 0
0 1
分析：比赛的时候做出来的队友说的是数论结论题，比赛后我是按照题目意思直接模拟A掉的。。
根据题目给出的数字p按照题目的意思我们可以得到一个2*p行，p列的矩阵
其中1<=i<=p,1<=j<=p时：mapn[i][j] = ((i-1)+(j-1))%p
p+1<=i<=2*p,1<=j<=p时：mapn[i][j] = ((i-1)*(j-1))%p
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = pow(2,10)+10;
const double eps = 1e-8;
const ll mod = 1e9 + 7;
const ll inf = 1e9;
const double pi = acos(-1.0);
ll mapn[2*maxn][maxn];
int main() {
ll T, p;
scanf("%lld",&T);
while(T--) {
memset(mapn,0,sizeof(mapn));
scanf("%lld",&p);
for( ll i = 1; i <= 2*p; i ++ ) {
for( ll j = 1; j <= p; j ++ ) {
if( i <= p ) {
mapn[i][j] = ((i-1)+(j-1))%p;
} else {
mapn[i][j] = (i-1)*(j-1)%p;
}
if( j != p ) {
printf("%lld ",mapn[i][j]);
} else {
printf("%lld\n",mapn[i][j]);
}
}
}
}
return 0;
}