网店代运营网站seo搜索引擎优化关键词
文章目录
- 前缀和
 - 一维前缀和
 - 公式
 - CODE
 
- 二维前缀和
 - 公式
 - CODE
 
- 差分
 - 一维差分
 - 思路
 - 作用
 - CODE
 
- 二维差分
 - 思路
 - CODE
 
前缀和
一维前缀和
板子题:https://www.acwing.com/activity/content/problem/content/829/
公式
S [ i ] = a [ i ] + S [ i − 1 ] S[i] = a[i] + S[i - 1] S[i]=a[i]+S[i−1]
CODE
#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 1e5 + 10;
int n, m, l, r;
int a[N], s[N];int main()
{cin >> n >> m;for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);s[i] = s[i - 1] + a[i];}while(m--){cin >> l >> r;printf("%d\n", s[r] - s[l - 1]);}
}
 
二维前缀和
板子题:https://www.acwing.com/activity/content/problem/content/830/
公式
S [ i ] [ j ] = S [ i − 1 ] [ j ] + S [ i ] [ j − 1 ] − S [ i − 1 ] [ j − 1 ] + a [ i ] [ j ] S[i][j] = S[i - 1][j] + S[i][j - 1] - S[i - 1][j - 1] + a[i][j] S[i][j]=S[i−1][j]+S[i][j−1]−S[i−1][j−1]+a[i][j]
CODE
#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 1010;
int n, m, q;
int x1, x2, y1, y2;
int a[N][N], s[N][N];int main()
{cin >> 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];}while (q -- ){cin >> x1 >> y1 >> x2 >> y2;printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);}
}
 
差分
一维差分
板子题:https://www.acwing.com/activity/content/problem/content/831/
思路
差分其实是前缀和的逆运算,我们假想有一个数组b[],它的前缀和是数组a[],也就是说:
  b [ i ] = a [ i ] − a [ i − 1 ] b[i] = a[i] - a[i - 1] b[i]=a[i]−a[i−1]
作用
这个b[]数组有什么用呢?
 在我们对a[]的元素进行加减操作时,如果采用遍历a[]的方法,时间是  o ( N ) o(N) o(N) 的,但是如果我们用b[]对其优化可以使时间复杂度降到  o ( 1 ) o(1) o(1)。
对a[]的  [ i , j ] [i, j] [i,j] 段进行+k操作,我们可以在 b[i] + k并在b[j + 1] - k。当我们对b[]求前缀和时,从i开始的每个元素都会+k,但是我们只要加到a[j]就结束了,所以在a[j + 1]进行归位。
CODE
#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 1e5 + 10;
int n, m;
int l, r, c;
int a[N], b[N];void insert(int l, int r, int c){b[l] += c;b[r + 1] -= c;
}int main()
{cin >> n >> m;for(int i = 1; i <= n; ++i){ scanf("%d", &a[i]);insert(i, i, a[i]);}while (m -- ){cin >> l >> r >> c;insert(l, r, c);}for(int i = 1; i <= n; ++i) printf("%d ", b[i] += b[i - 1]);
}
 
整个差分数组的精髓就在于insert()函数,非常巧妙啊,尤其是在读入阶段对b[]数组进行初始化时的操作,这个操作的意义如下:

 来源:https://www.acwing.com/activity/content/code/content/39799/
二维差分
板子题:https://www.acwing.com/activity/content/problem/content/832/
思路
答题思路跟一维差分差不多,借鉴二维前缀和的操作我们可以得到以下公式:
  a [ i ] [ j ] = b [ i ] [ j ] − b [ i − 1 ] [ j ] − b [ i ] [ j − 1 ] + b [ i − 1 ] [ j − 1 ] a[i][j] = b[i][j] - b[i - 1][j] - b[i][j - 1] + b[i - 1][j - 1] a[i][j]=b[i][j]−b[i−1][j]−b[i][j−1]+b[i−1][j−1]
那我们插入函数该怎么写呢?
 一样的原理:
  b [ x 1 ] [ y 1 ] + = c b [ x 2 + 1 ] [ y 1 ] − = c b [ x 1 ] [ y 2 + 1 ] − = c b [ x 2 + 1 ] [ y 2 + 1 ] + = c b[x1][y1] += c\\ b[x2 + 1][y1] -= c\\ b[x1][y2 + 1] -=c\\ b[x2 + 1][y2 + 1] += c b[x1][y1]+=cb[x2+1][y1]−=cb[x1][y2+1]−=cb[x2+1][y2+1]+=c
CODE
#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;int n, m, q;
const int N = 1010;
int a[N][N], b[N][N];
int x1, y1, x2, y2, c;void insert(int x1, int y1, int x2, int y2, int c){b[x1][y1] += c;b[x2 + 1][y1] -= c;b[x1][y2 + 1] -= c;b[x2 + 1][y2 + 1] += c;
}int main()
{cin >> n >> m >> q;for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j){scanf("%d", &a[i][j]);insert(i, j, i, j, a[i][j]);}while(q--){cin >> x1 >> y1 >> x2 >> y2 >> c;insert(x1, y1, x2, y2, c);}for(int i = 1; i <= n; ++i){for(int j = 1; j <= m; ++j){printf("%d ", b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1]);}printf("\n");   }
}
