1.2 基础算法(二)
这是我的一个算法网课学习记录,道阻且长,好好努力
1.2.1 高精度
A + B len 106
A - B len 106
A * a len(A) <= 106, a <= 1e4;
A / a
大整数 big decimal
大整数存储 (倒着来,第一位存储各位;如果要进位,在最后补上一个数据;存储格式是一样的)
-
高精度加法
逢10进1,Ci = Ai + Bi + t, (t = 0, 1);
// 高精度加法
#include <iostream>
#include <vector> //size
using namespace std;
// const int N = 1e6 + 10;
// C = A + B
vector<int> add(vector<int> &A, vector<int> &B) // 加&,使其不copy,提高效率
{
vector<int> C;
int t = 0; // 进位
for (int i = 0; i < A.size() || i < B.size(); i ++ )
{
if (i < A.size()) t += A[i];
if (i < B.size()) t += B[i];
C.push_back(t % 10);
t /= 10;
}
if (t) C.push_back(1);
return C;
}
int main()
{
string a, b;
vector<int> A, B;
cin >> a >> b; // a = "123456"
// 倒序存储
// A = [6, 5, 4, 3, 2, 1]
for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');
auto C = add(A, B); // auto用于自动型别推导
for (int i = C.size() - 1; i >= 0; i -- ) printf("%d", C[i]);
return 0;
}
- 高精度减法
// 高精度减法
#include <iostream>
#include <vector> //size
using namespace std;
// const int N = 1e6 + 10;
bool cmp(vector<int> &A, vector<int> &B)
{
if (A.size() != B.size()) return A.size() > B.size();
for (int i = A.size() - 1; i >= 0; i -- )
if (A[i] != B[i])
return A[i] > B[i];
return true;
}
// C = A - B
vector<int> sub(vector<int> &A, vector<int> &B) // 加&,使其不copy,提高效率
{
vector<int> C;
int t = 0; // 借位与否
for (int i = 0; i < A.size(); i ++ )
{
t = A[i] - t;
if (i < B.size()) t -= B[i];
C.push_back((t + 10) % 10);
if (t < 0) t = 1;
else t = 0;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main()
{
string a, b;
vector<int> A, B;
cin >> a >> b; // a = "123456"
// 倒序存储
// A = [6, 5, 4, 3, 2, 1]
for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');
int main()
{
string a, b;
vector<int> A, B;
cin >> a >> b; // a = "123456"
// 倒序存储
// A = [6, 5, 4, 3, 2, 1]
for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');
if (cmp(A, B))
{
auto C = sub(A, B); // auto用于自动型别推导
for (int i = C.size() - 1; i >= 0; i -- ) printf("%d", C[i]);
}
else
{
auto C = sub(B, A);
printf("-");
for (int i = C.size() - 1; i >= 0; i -- ) printf("%d", C[i]);
}
return 0;
}
return 0;
}
- 高精度乘法
#include <iostream>
#include <vector>
using namespace std;
// C = A * b
vector<int> mul(vector<int> &A, int b)
{
vector<int> C;
int t = 0; // 进位
for (int i = 0; i < A.size() || t; i ++ )
{
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
return C;
}
int main()
{
string a;
int b;
cin >> a >> b;
vector<int> A;
for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
auto C = mul(A, b);
for (int i = C.size() - 1; i >= 0; i -- ) printf("%d", C[i]);
return 0;
}
- 高精度除法
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// 返回商,并更新r为余数
vector<int> div(vector<int> &A, int b, int &r) {
vector<int> C;
r = 0;
// A是将一个整数逆序存储的,所以其最高位是最后一位,要以从最高位到最低位的次序计算
for (int i = A.size() - 1; i >= 0; i--) {
r = r * 10 + A[i];
C.push_back(r / b);
r %= b;
}
reverse(C.begin(), C.end());
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main() {
string a;
int b;
cin >> a >> b;
vector<int> A;
for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
int r;
auto C = div(A, b, r);
for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
cout << endl << r << endl;
return 0;
}
1.2.2 前缀和
(其实这是一个思想)
- 一维前缀和
原i: a1 ,a2, a3, … , an
前缀和: Si = a1 + a2 + a3 + … + ai
两段相减可以得到任意的一段数组的和
Sr - SL-1 ---->O(1)
// 一维前缀和
#include <iostream>
using namespace std;
const int N = 1e6 + 10;
int n, m;
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]; //前缀和的初始化
while (m -- )
{
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", s[r] - s[l - 1]); // 区间和的计算
}
return 0;
}
关于初始化int数组,记住三点:
- 全局数组,未初始化时,默认值都是 0;
- 局部数组,未初始化时,默认值为随机的不确定的值;
- 局部数组,初始化一部分时,未初始化的部分默认值为 0;
-
二维前缀和
主要是公式。具有容斥原理的思想;重复step by step。
#include <iostream>
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 <= n; 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, x2, y1, y2;
scanf("%d%d%d%d", &x1, &x2, &y1, &y2);
printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]; // 求子矩阵的和(靠近右下角边框)
}
return 0;
}
1.2.3 差分
差分是前缀和的逆运算
并不需要考虑如何构造,需要考虑如何更新
- 一维差分
#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]); // 插入
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];
printf("%d ", b[i]);
}
return 0;
}
for (int i = 1; i <= n; i ++ ) insert(i, i, a[i]);
这一步将辅助数组b转换成了差分的形式:
a1 a2 a3 a4 … an-1 an
-a1 -a2 -a3… -an-2 -an-1 -an
- 二维差分
// 差分矩阵
#include <iostream>
using namespace std;
const int N = 1e2 + 10;
int n, m, q;
int a[N][N], b[N][N];
void insert(int x1, int y1, int x2, int y2)
{
b[x1][y1] += c;
b[x2 + 1][y1] -= c;
b[x1][y2 + 1] -= c;
b[x2 + 1][y2 + 1] += c;
}
// 对于矩形的左上角顶点运用容斥原理
int main()
{
scanf("%d%d%d", &n, &m, &q);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= n; j ++ )
scanf("%d", &a[i][j]);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= n; 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 <= n; 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 <= n; j ++ )
printf("%d", b[i][j]);
puts("");
}
return 0;
}