一维/二维_差分/前缀和_代码实现

一维前缀和

代码实现：

• 计算公式：S[r]-S[l-1]=l~r范围和

#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <stdio.h>

using namespace std;

const int N=100010;

int n, m;					//n表示数组已存入的数个数,m表示需要查询的次数
int A[N], S[N];

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++)S[i] = S[i - 1] + A[i];

	while (m--)
	{
		int l, r;
		scanf("%d%d", &l, &r);				//区间和的计算
		printf("%d\n",S[r] - S[l - 1]);
	}
	system("pause");
	return 0;
}

一维前缀和(2)

//计算一维前缀和(q次询问，输入l和r，输出l->r之间的和值)
//公式：sum[l->r]=sum[r]-sum[l-1];sum[i]=a[i]+sum[i-1];

void test01()
{
	int a[1000] = { 0 }, sum[1000] = { 0 }; sum[0] = 0;
	int n, q, l, r;
    
	cin >> n >> q;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
		sum[i] = sum[i - 1] + a[i];
	}
	while (q--) {
		cin >> l >> r;
		printf("%d", sum[r] - sum[l - 1]);
	}
}

二维前缀和

代码实现：

#define _CRT_SECURE_NO_WARNINGS 1
#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]);

	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; 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[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);		//求子矩阵
	}
	return 0;
}

二维前缀和(2)

//计算二维前缀和		
//前缀和数组公式：sum[x][y] = sum[x - 1][y] + sum[x][y - 1] - sum[x - 1][y - 1] + a[x][y];
//子矩阵	    ：sum[x_2-x_1][y_2-y_1]=sum[x_2][y_2]-sum[x_2][y_1-1]-sum[x_1-1][y_2]+sum[x_1-1][y_1-1];
//例题：求区间[x_1,y_1]->[x_2,y_2]区间中数值之和
void test03()
{
	int a[50][50] = { 0 }, sum[50][50] = { 0 };
	int n, m, q;			//n行m列,q次询问

	cin >> n >> m >> q;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
		{
			cin >> a[i][j];
			sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + a[i][j];
		}
	while (q--)
	{
		int x_1, y_1, x_2, y_2;
		cin >> x_1 >> y_1 >> x_2 >> y_2;
		cout << sum[x_2][y_2] - sum[x_1 - 1][y_2] - sum[x_2][y_1 - 1] + sum[x_1 - 1][y_1 - 1] << endl;
	}
}

一维差分

• 原理：有一数列{a1,a2,a3…an}构造{b1,b2,b3…bn}使得an=b1+b2+…+bn.
• 代码实现：

给区间[l, r]中的每个数加上c：B[l] += c, B[r + 1] -= c

#include <iostream>
using namespace std;

const int N = 10010;

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]);
	while (m--)
	{
		int l, r, c;
		scanf("%d%d%d", &l, &r, &c);
		insert(l, r, c);
	}
	for (int i = 1; i <= n; i++)b[i] += b[i - 1];
	for (int i = 1; i <= n; i++)printf("%d", b[i]);
	system("pause");
	return 0;
}

一维差分(2)

//计算一维差分
//例题：在n个数中，进行q次操作，每次询问一个区间[l->r]，使得区间中所有数加上一个数值c
//由已知数值序列A{a_1,a_2,a_3.....a_n}构造差分序列D{b_1,b_2,b_3.....b_n}使得a_n=b_1+b_2+.....+b_n;4
//公式：A[L->R]中每个数值+c    <=>    D[L]+c,D[R+1]-c
int  D[1000] = { 0 };
void insert(int l, int r, int c) { D[l] += c, D[r + 1] -= c; }
void test02()
{
	int A[1000] = { 0 };
	int n, q, l, r, c;

	cin >> n >> q;
	for (int i = 1; i <= n; i++)//创建数值序列A并构造A的差分序列D
	{
		cin >> A[i]; 
		insert(i, i, A[i]);
	}
	while (q--)
	{
		cin >> l >> r >> c;
		insert(l, r, c);
	}
	for (int i = 1; i <= n; i++)
	{
		D[i] += D[i - 1];
		cout << D[i] << " ";
	}//将差分数组D做一次前缀求和就行
}

二维差分

代码实现：

给以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵中的所有元素加上c：
S[x1, y1] += c, S[x2 + 1, y1] -= c, S[x1, y2 + 1] -= c, S[x2 + 1, y2 + 1] += c
    
#define _CRT_SECURE_NO_WARNINGS 1
#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[x1][y2 + 1] -= c;
	b[x2 + 1][y1] -= 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]);

	return 0;
}

二维差分(2)

//计算二维差分
/* 例题：在一个 n* m 的矩阵中进行q次操作，将(x_1, y_1)->(x_2, y_2)中的元素都加上数值c
		 即给以(x_1, y_1)为左上角，(x_2, y_2)为右下角的子矩阵中的所有元素加上c          */
//由已知矩阵创建差分矩阵，原矩阵每个对应元素A(i,j)的值都等于以差分矩阵D(0,0)为起点D(i,j)为终点的子矩阵所有元素的值求和
//公式：D[x1][y1] += c; D[x1][y2 + 1] -= c; D[x2 + 1][y1] -= c; D[x2 + 1][y2 + 1] += c;
int D[100][100];
void insert(int x_1, int y_1, int x_2, int y_2,int c) 
{ 
	D[x_1][y_1] += c;
	D[x_1][y_2 + 1] -= c;
	D[x_2 + 1][y_1] -= c;
	D[x_2 + 1][y_2 + 1] += c;
}
void test04()
{
	int n, m, q;
	int A[100][100] = { 0 };
	cin >> n >> m >> q;
	for(int x=1;x<=n;x++)
		for (int y = 1; y <= m; y++)
		{
			cin >> A[x][y];
			insert(x, y, x, y, A[x][y]);
		}
	while (q--)
	{
		int x_1, x_2, y_1, y_2, c;
		cin >> x_1 >> y_1 >> x_2 >> y_2 >> c;
		insert(x_1, y_1, x_2, y_2, c);
	}
	for (int x = 1; x <= n; x++)
	{
		for (int y = 1; y <= m; y++)
		{
			D[x][y] += D[x - 1][y] + D[x][y - 1] - D[x - 1][y - 1];
			cout << D[x][y] << " ";
		}
		cout << endl;
	}
	return;
}