c++算法学习笔记 (5)前缀和+差分

#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int n, m;
int a[N], s[N];
int main()
{
  // ios::sync_with_stdio(false); // 提高cin的读取速度，但不能使用scanf
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]);
  for (int i = 0; i <= n; i++)
    s[i] = s[i - 1] + a[i];
  while (m--)
  {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", s[r] - s[l - 1]);
  }
  return 0;
}

#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q;
int a[N][N], s[N][N];
int main()
{
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++)
  {
    for (int j = 1; j <= m; j++)
    {
      scanf("%d", &a[i][j]);
      s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j]; // 求前缀和
    }
  }
  while (q--)
  {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    printf("%d\n", s[x2][y2] - s[x2][y1 - 1] - s[x1 - 1][y2] + s[x1 - 1][y1 - 1]); // 算子矩阵的和
  }
  return 0;
}
 

#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int n, m;
int a[N], b[N];
int main()
{
  scanf("%d%d", &n, &m);
  a[0] = 0;
  for (int i = 1; i <= n; i++)
  {
    scanf("%d", &a[i]);
    b[i] = a[i] - a[i - 1]; // 差分
  }
  while (m--)
  {
    int l, r, c;
    scanf("%d%d%d", &l, &r, &c);
    b[l] += c;
    b[r + 1] -= c;
  }
  b[0] = 0;
  for (int i = 1; i <= n; i++)
  {
    a[i] = a[i - 1] + b[i];
  }
  for (int i = 1; i <= n; i++)
  {
    printf("%d ", a[i]);
  }
  return 0;
}

// 差分
#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q;
int a[N][N], b[N][N];
int main()
{
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++)
  {
    for (int j = 1; j <= m; j++)
    {
 
      scanf("%d", &a[i][j]);
      b[i][j] = a[i][j] - a[i - 1][j] - a[i][j - 1] + a[i - 1][j - 1]; // 差分
    }
  }
  while (q--)
  {
    int x1, y1, x2, y2, c;
    scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &c);
    b[x1][y1] += c;
    b[x2 + 1][y1] -= c;
    b[x1][y2 + 1] -= c;
    b[x2 + 1][y2 + 1] += c;
  }
  for (int i = 1; i <= n; i++)
  {
    for (int j = 1; j <= m; j++)
    {
      a[i][j] = a[i-1][j]+a[i][j-1]-a[i-1][j-1]+b[i][j];
    }
  }
  for (int i = 1; i <= n; i++)
  {
    for (int j = 1; j <= m; j++)
    {
      printf("%d ", a[i][j]);
    }
    printf("\n");
  }
  return 0;
}