CF621E Wet Shark and Blocks

Posted 2021-10-28 Ariel_

CF621E

题面

solution

一个 显然的 \\(dp\\)

设 \\(f_{i, j}\\) 表示考虑了前 \\(i\\) 个格子，当前余数为 \\(j\\) 的方案数。

转移就枚举当前选的数

\\[f_{i + 1, (j * 10 + v) \\% x} += f_{i, j} \\]

然后你会发现转移方程的每个 \\(f_{i}\\) 的转移都是一样的。

所以可以用矩阵加速转移。

初始状态：

\\(f_0 = \\{1, 0, 0, \\dots 0\\}\\)

对于每个数 \\(v\\) 修改矩阵：

\\(base[j][(j * 10 + v)\\% x] += 1\\)

最后答案就是 \\(base^b \\times f[0]\\) 的第 \\(k\\) 行。

code

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int MAXA = 1e4 + 5;
const int MAXB = 1e5 + 5;
const int MAXC = 1e6 + 5;
const int mod = 1e9 + 7;
int read() {
  int x = 0, f = 1; char c = getchar();
  while(c < \'0\' || c > \'9\') {if(c == \'-\') f = -1;c = getchar();}
  while(c >= \'0\' && c <= \'9\') {x = x * 10 + c - \'0\';c = getchar();}
  return x * f;
}
int n, b, k, x;
struct Matrix{
   int a[110][110];
   Matrix() {memset(a, 0, sizeof a);}
   Matrix operator * (const Matrix &rhs)const{
       Matrix ret;
       for (int i = 0; i < x; i++) 
       	 for (int j = 0; j < x; j++) 
		   for (int k = 0; k < x; k++) 
			ret.a[i][j] = (ret.a[i][j] + a[i][k] * rhs.a[k][j] % mod) % mod;  
      return ret;
   }
}base, Ans;
Matrix pow(int k) {
   Matrix ret;
   for (int i = 0; i < x; i++) ret.a[i][i] = 1;
   while(k) {
   	 if(k & 1) ret = ret * base;
   	 k >>= 1;
   	 base = base * base;
   }
   return ret;
}
signed main() {
   n = read(), b = read(), k = read(), x = read();
   for (int i = 1; i <= n; i++) {
   	  int val = read();
   	  for (int j = 0; j < x; j++) 
   	     base.a[j][(j * 10 % x + val % x) % x]++;	  
   }
   Ans = pow(b);
   cout<<Ans.a[0][k];
   return 0;
}

Wet Shark and Bishops（思维）

Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 to 1000. Rows are numbered from top to bottom, while columns are numbered from left to right.

Wet Shark thinks that two bishops attack each other if they share the same diagonal. Note, that this is the only criteria, so two bishops may attack each other (according to Wet Shark) even if there is another bishop located between them. Now Wet Shark wants to count the number of pairs of bishops that attack each other.

Input

The first line of the input contains n (1 ≤ n ≤ 200 000) — the number of bishops.

Each of next n lines contains two space separated integers x_i and y_i (1 ≤ x_i, y_i ≤ 1000) — the number of row and the number of column where i-th bishop is positioned. It‘s guaranteed that no two bishops share the same position.

Output

Output one integer — the number of pairs of bishops which attack each other.

Example

Input

5
1 1
1 5
3 3
5 1
5 5

Output

6

Input

3
1 1
2 3
3 5

Output

0

Note

In the first sample following pairs of bishops attack each other: (1, 3), (1, 5), (2, 3), (2, 4), (3, 4) and (3, 5). Pairs (1, 2), (1, 4), (2, 5) and (4, 5) do not attack each other because they do not share the same diagonal.

 

 

题意：在对角线上的相会相互攻击，求出能够相互攻击的相的对数？

题解：仔细观察能够发现，在同一对角线上的相满足x+y or x-y相等，这里需要注意的是x-y可能会出现负数，所以这里用x-y+3000来代替x-y

AC代码

#include<stdio.h>
#include<string.h>

int main()
{
    int n;
    int x, y;
    int attack[5000];
    int sum;
    while(~scanf("%d", &n))
    {
        sum = 0;
        memset(attack, 0, sizeof(attack));
        for(int i = 0; i < n; i++)
        {
            scanf("%d %d", &x, &y);
            attack[x+y]++;
            attack[x-y+3000]++;
        }
        for(int i = 0; i < 5000; i++)
        {
            if(attack[i] > 1)
                sum += attack[i]*(attack[i]-1)/2;
        }
        printf("%d\n", sum);
    }

    return 0;
}

View Code