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1024 矩阵中不重复的元素

1 秒
131,072 KB
10 分
2 级题

一个m*n的矩阵。该矩阵的第一列是a^b,(a+1)^b,.....(a + n - 1)^b 第二列是a^(b+1),(a+1)^(b+1),.....(a + n - 1)^(b+1) ....... 第m列是a^(b + m - 1),(a+1)^(b + m - 1),.....(a + n - 1)^(b + m - 1) (a^b表示a的b次方）下面是一个4*4的矩阵： 2^2=4, 2^3=8, 2^4=16, 2^5=32 3^2=9, 3^3=27, 3^4=81, 3^5=243 4^2=16, 4^3=64, 4^4=256, 4^5=1024 5^2=25, 5^3=125, 5^4=625, 5^5=3125 问这个矩阵里有多少不重复的数（比如4^3 = 8^2，这样的话就有重复了) 2^2=4, 2^3=8, 2^4=16, 2^5=32 3^2=9, 3^3=27, 3^4=81, 3^5=243 4^2=16, 4^3=64, 4^4=256, 4^5=1024 m = 4, n = 3, a = 2, b = 2。其中2^4与4^2是重复的元素。

输入

输入数据包括4个数：m,n,a,b。中间用空格分隔。m，n为矩阵的长和宽（2 <= m,n <= 100)。a，b为矩阵的第1个元素，a^b（2 <= a , b <= 100）。

输出

输出不重复元素的数量。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 20005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;

inline int rd() {
  int x = 0;
  char c = getchar();
  bool f = false;
  while (!isdigit(c)) {
    if (c == '-') f = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}


ll gcd(ll a, ll b) {
  return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
  if (!b) {
    x = 1; y = 0; return a;
  }
  ans = exgcd(b, a%b, x, y);
  ll t = x; x = y; y = t - a / b * y;
  return ans;
}
*/

double A[102][102];
int  n, m;
int a, b;
setst;

int main()
{
  //  ios::sync_with_stdio(0);
  m = rd(); n = rd();
  a = rd(); b = rd();
  int ans = 0;
  for (int j = 1; j <= m; j++) {
    for (int i = 1; i <= n; i++) {
      A[i][j] = 1.0*b * log2(a + i - 1);
      st.insert(A[i][j]);
    }
    b++;
  }
  ans = st.size();
  printf("%d\n", ans);
  return 0;
}

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