增加大整数运算的示例实现

SWBitOp.cpp

@@ -36,4 +36,74 @@ long long SWBitOp::Mutiple(long long a, long long b, long long p)
 		b >>= 1;
 	}
 	return res;
-}
+}
+
+vector<int> SWBitOp::add(vector<int>& l, vector<int>& r)
+{
+	vector<int> res;
+	int bit = 0;
+	for (int i = 0; i < l.size() || i < r.size() || bit; i++)
+	{
+		if (i < l.size()) bit += l[i];
+		if (i < r.size()) bit += r[i];
+		res.push_back(bit % 10);
+		bit /= 10;
+	}
+
+	return res;
+}
+
+//besure content of l is bigger than content of r
+vector<int> SWBitOp::sub(vector<int>& l, vector<int>& r)
+{
+	vector<int> res;
+	int bit = 0;
+	for (int i = 0; i < l.size() || bit; i++)
+	{
+		bit = l[i] - bit;
+		if (i < r.size()) bit -= r[i];
+		res.push_back((bit + 10) % 10);
+		if (bit < 0)
+			bit = 1;
+		else
+			bit = 0;
+	}
+	while (res.size() > 1 && res.back() == 0) res.pop_back();
+
+	return res;
+}
+
+vector<int> SWBitOp::multi(vector<int>& l, int r)
+{
+	vector<int> res;
+	int bit = 0;
+	for (int i = 0; i < l.size() || bit; i++)
+	{
+		if(i < l.size()) bit += l[i] * r;
+		res.push_back(bit % 10);
+		bit /= 10;
+	}
+	while (res.size() > 1 && res.back() == 0) res.pop_back();
+	return res;
+}
+
+vector<int> SWBitOp::div(vector<int>& l, int r,int &rem)
+{
+	vector<int> res;
+	int bit = 0;
+	for (int i = l.size() - 1; i >= 0; i--)
+	{
+		if (i >= 0)
+			bit = bit * 10 + l[i];
+		res.push_back(bit / r);
+		bit %= r;
+	}
+	reverse(res.begin(),res.end());
+
+	while (res.size() > 1 && res.back() == 0) res.pop_back();
+
+	rem = bit;
+	return res;
+}
+
+

SWBitOp.h

@@ -1,4 +1,6 @@
 #pragma once
+#include <vector>
+using namespace std;
 class SWBitOp
 {
 public:
@@ -10,5 +12,13 @@ class SWBitOp
 
 	//#2 a * b % p µÄÖµ(64bit)
 	static long long Mutiple(long long a, long long b, long long p);
+
+	static vector<int> add(vector<int> & l,vector<int>& r);
+
+	static vector<int> sub(vector<int>& l, vector<int>& r);
+
+	static vector<int> multi(vector<int> & l,int r);
+
+	static vector<int> div(vector<int> & l,int r,int& rem);
 };
 

Test.cpp

@@ -1,8 +1,11 @@
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
+#include <iostream>
+using namespace std;
 #include "src/Algo.h"
 #include "CClock.h"
+#include "SWBitOp.h"
 
 void testQickSort(int src[],int n)
 {
@@ -67,5 +70,69 @@ void testRangeOfNumber(int src[], int n,int k)
 
 	clock.end();
 
+	return;
+}
+
+void testAdd(vector<int>& l, vector<int>& r)
+{
+	CClock clock;
+
+	auto res =  SWBitOp::add(l,r);
+
+	for (int i = res.size() - 1; i >= 0; i--)
+		cout << res[i];
+
+	cout << endl;
+
+	clock.end();
+
+	return;
+}
+
+void testSub(vector<int>& l, vector<int>& r)
+{
+	CClock clock;
+
+	auto res = SWBitOp::sub(l, r);
+
+	for (int i = res.size() - 1; i >= 0; i--)
+		cout << res[i];
+
+	cout << endl;
+
+	clock.end();
+
+	return;
+}
+
+void testMulti(vector<int>& l, int r)
+{
+	CClock clock;
+
+	auto res = SWBitOp::multi(l, r);
+
+	for (int i = res.size() - 1; i >= 0; i--)
+		cout << res[i];
+
+	cout << endl;
+
+	clock.end();
+
+	return;
+}
+
+void testDiv(vector<int>& l, int r,int & rem)
+{
+	CClock clock;
+
+	auto res = SWBitOp::div(l, r,rem);
+
+	for (int i = res.size() - 1; i >= 0; i--)
+		cout << res[i];
+
+	cout << endl << rem << endl;
+
+	clock.end();
+
 	return;
 }

main.cpp

@@ -1,4 +1,5 @@
 #include <iostream>
+#include <vector>
 #include "MinStack.h"
 using namespace std;
 
@@ -8,6 +9,10 @@ extern void testKMins(int src[], int n, int k);
 extern void testMergeSort(int src[], int n, int tmp[]);
 extern void testReversePairNum(int src[], int n, int tmp[], long long* pairs);
 extern void testRangeOfNumber(int src[], int n,int k);
+extern void testAdd(vector<int>& l, vector<int>& r);
+extern void testSub(vector<int>& l, vector<int>& r); 
+extern void testMulti(vector<int>& l, int r);
+extern void testDiv(vector<int>& l, int r, int& rem);
 
 int main()
 {
@@ -20,6 +25,9 @@ int main()
 
 	long long pairs = 0;
 
+	vector<int> l = {2,2,1};
+	vector<int> r = {3,2,1};
+
 	//testQickSort(f,sizeof(f) / sizeof(f[0]));
 
 	//testKMins(f,sizeof(f) / sizeof(f[0]),4);
@@ -28,7 +36,16 @@ int main()
 
 	//testReversePairNum(f, sizeof(f) / sizeof(f[0]), tmp, &pairs);
 
-	testRangeOfNumber(f, sizeof(f) / sizeof(f[0]),2);
+	//testRangeOfNumber(f, sizeof(f) / sizeof(f[0]),2);
+
+	testAdd(l,r);
+
+	testSub(r,l);
+
+	testMulti(l,30);
+
+	int remain = 0;
+	testDiv(l,3,remain);
 
 	delete [] tmp;
 }

算法基础点.txt

@@ -16,4 +16,22 @@
 	有单调性一定可以二分，二分的场景不一定需要有单调性
 	本质是用来找边界，这个边界在左半边区间是满足的，在右半边区间是不满足的。因是整数区间，所以两个区间没有交点
 对应两个点：最后一个满足条件1的点，第一个满足条件2的点
-	每次选择时，选择答案所在的区间，边界点不丢失
+	每次选择时，选择答案所在的区间，边界点不丢失
+
+4 大整数运算
+	大整数的每一位都保存到数组中，数组的低位保存数的低位，在进位时，在数据最后一位加1就行，不用移动整个数组
+	数的表示及存储
+	A3 A2 A1 A0
++	   B2 B1 B0
+
+	A3 A2 A1 A0
+-	   B2 B1 B0
+	进位，0或者1
+
+	除法中余数
+
+5 前缀和数组 
+	一维 sum[i] = sum[i - 1] + A[i]; i从1开始，避免边界条件 快速求出一段区间的和
+	二维 快速求出子矩阵的和
+
+	差分数组的构造