试题背景
问题描述
输入格式
输出格式
样例
输入
4 16 1 6
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
输出
7
C++求解
暴力方法
#include<iostream>
using namespace std;
int main()
{
int n,L,r,t;
int result=0;
cin>>n>>L>>r>>t;
int A[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>A[i][j];
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
//i j
int sum=0;
int num=0;
for(int k=i-r;k<=i+r;k++){
if(k>=0&&k<n){
for(int p=j-r;p<=j+r;p++){
if(p>=0&&p<n){
sum+=A[k][p];
num++;
}
}
}
}
double avg=(double)sum/num;
if(avg<=t) result++;
}
}
cout<<result;
return 0;
}
一些聪明的方法
先求出每一列第一个对应数的sum和num,在此基础上进行加减求这一行的数的sum和num
#include <iostream>
using namespace std;
int main()
{
int n, L, r, t;
int matrix[600][600];
int res = 0;
int temp;
cin >> n >> L >> r >> t;
//input
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
cin >> temp;
matrix[i][j] = temp;
}
}
//data process
int sum = 0, num = 0, lastsum = 0, lastnum = 0;
for(int i = 0; i < n; i++)//先求第一列
{
sum = num = 0;
int bound_x = i + r > n - 1? n - 1 : i + r;
int bound_y = 0 + r > n - 1? n - 1 : 0 + r;
int start_x = i - r > 0 ? i - r : 0;
for(int k = start_x; k <= bound_x; k++)
for(int m = 0; m <= bound_y; m++)//第一排的y都从0开始
{
num++;
sum += matrix[k][m];
}
double avg = (double)sum / num;
lastnum = num; lastsum = sum;
if(avg <= t)
res++;
for(int j = 1; j < n; j++)//每一列对应的排
{
sum = lastsum;
if(j - r <= 0)
{
num = lastnum + bound_x - start_x + 1;
for(int m = start_x; m <= bound_x; m++)
sum += matrix[m][j+r];
avg = (double)sum / num;
}
else if(j + r > n - 1)
{
num = lastnum - (bound_x - start_x + 1);
for(int m = start_x; m <= bound_x; m++)
sum -= matrix[m][j - r - 1];
avg = (double)sum / num;
}
else
{
num = lastnum;
for(int m = start_x; m <= bound_x; m++)
sum += matrix[m][j+r] - matrix[m][j - r - 1];
avg = (double)sum / num;
}
if(avg <= t)
res++;
lastnum = num; lastsum = sum;//替换一下
}
}
cout << res;
return 0;
}