Use mathematics to explain the relationship between difference and prefix sum
Before learning the difference, you need to review the concept and definition of the prefix sum: click here to see the prefix and concept
We know the implementation of the prefix and code:s[ i ] = s[ i - 1] + a[ i ];(Equation 1)
That is, the array S stores the sum of the first i items of the array a; I believe this is easy to understand, but when it comes to the definition of difference, it is easy to confuse;
why is the difference in the inverse operation of the prefix sum?
Let's look at the definition of difference: difference is the difference between two adjacent numbers .
We can write directly according to its definition literally:s[ i ] = a[i] - a[i-1];(Equation 2)
That is, use the array S to record the difference between each adjacent number of the array a.
We observe these two formulas and reverse the array s and the array a in the prefix sum to obtain formula 2.
That is, in mathematics, let s = a, a = s, then a[i] = a[i-1] + s[i], put a[i-1] on the other side to become
s[1] = a[i]-a[i-1]. It can be seen that the prefix sum and difference are a kind of inverse operation.
We construct a set bn to satisfy:
av = b1 + b2 + b3 +… + bv;
construct bn so that the array n is called the prefix sum of the b array, and the b array is the difference of the a array
We can use difference to better solve the problem of interval sum:
given an array a, if we want to add a number c in one of the intervals [l, r], we may use for(int i = l ;i <= r; i++) a[i] += c;this time complexity for the first time It is O(n). If the interval [l,r] is very large, it is easy to process overtime. Therefore, we use differential operations s[l] += c; s[r + 1] -= c;to ensure that only c is added between the interval l and r.
Finally, we also need to restore the array a after adding c, directly "reverse", let's look atEquation 2, The way to get a[i], namelya[i] = s[i] + a[i -1]
Example: Difference
Enter a sequence of integers of length n.
Next, enter m operations, each operation contains three integers l, r, c, which means adding c to each number between [l, r] in the sequence.
Please output the sequence after all operations have been performed.
Input format The
first line contains two integers n and m.
The second line contains n integers, representing a sequence of integers.
The next m lines, each line contains three integers l, r, c, representing an operation.
The output format is
one line, containing n integers, representing the final sequence.
Data range
1≤n, m≤100000,
1≤l≤r≤n,
-1000≤c≤1000,
-1000≤The value of the element in the integer sequence ≤1000
Input example:
6 3
1 2 2 1 2 1
1 3 1
3 5 1
1 6 1
Sample output:
3 4 5 3 4 2
C++ implementation:
#include<iostream>
using namespace std;
const int N = 100010;
int n,m;
int a[N],b[N];
void insert(int l, int r,int c)
{
b[l] += c;
b[r + 1] -=c;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i = 1; i <= n;i ++ ) scanf("%d",&a[i]);
for(int i = 1; i <= n;i ++ ) insert(i, i ,a[i]);
//相当于 for(int i = 1; i <= n; i ++) b[i] = a[i] - a[i-1];
while(m -- )
{
int l, r, c;
scanf("%d%d%d",&l,&r,&c);
insert(l,r,c);
}
//求原数组差分后的情况：
for(int i = 1; i <= n; i ++) b[i] += b[i - 1]; //通过b[i] = b[i] + b[i-1]分解出原数组的a[]的情况
for(int i = 1; i <= n;i ++ ) printf("%d ", b[i]);
//或者 for(int i = 1; i <= n ; i ++) a[i] = b[i]+ a[i-1];
// for(int i = 1; i <= n ; i++) printf("%d ",a[i] );
return 0;
}
Differential matrix
input an integer matrix with n rows and m columns, and then input q operations, each operation contains five integers x1, y1, x2, y2, c, where (x1, y1) and (x2, y2) represent a sub The coordinates of the upper left corner and the lower right corner of the matrix.
Each operation must add c to the value of each element in the selected sub-matrix.
Please output the matrix after all operations are completed.
Input format The
first line contains integers n, m, q.
The next n rows, each row contains m integers, representing a matrix of integers.
The next q lines, each line contains 5 integers x1, y1, x2, y2, c, representing an operation.
The output format
has n rows and m integers in each row, representing the final matrix after all operations are completed.
Data range
1≤n, m≤1000,
1≤q≤100000,
1≤x1≤x2≤n,
1≤y1≤y2≤m,
−1000≤c≤1000,
−1000≤The value of the element in the matrix≤1000
input Example:
3 4 3
1 2 2 1
3 2 2 1
1 1 1 1
1 1 2 2 1
1 3 2 3 2
3 1 3 4 1
Sample output:
2 3 4 1
4 3 4 1
2 2 2 2
#include<iostream>
using namespace std;
const int N = 1010;
int n, m, q;
int a[N][N], b[N][N];
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()
{
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]);
}
for(int i= 1; i <= n ;i ++)
{
for(int j = 1; j <= m ; j ++)
{
insert(i, j, i, j, a[i][j]);
}
}
while(q--)
{
int x1, y1, x2, y2, c;
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 ++)
b[i][j] += b[i - 1][j] + b[i][j -1] - b[i -1][j - 1];
}
for(int i = 1; i <= n; i ++)
{
for(int j = 1; j <= m; j ++)
printf("%d ",b[i][j]);
puts(" ");
}
return 0;
}