1. 大整数存储
   数组存储，个位在前，有助于进位运算

例如，高精度加法

#include <iostream>
#include <vector>

using namespace std;

vector<int> add(vector<int> &A, vector<int> &B)
{
	vector<int> C;
	int t = 0;
	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;
	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);
	
	for(int i = C.size() -1;i>=0;i--) printf("%d",C[i]);
	
	return 0;
 } 

高精度减法模板

#include <iostream>
#include <vector>

using namespace std;

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; 
} 

vector<int> sub(vector<int> &A,vector<int> &B)
{
	vector<int> C;
	for(int i = 0,t=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;
	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);
		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;
 } 

高精度乘法模板

#include <iostream>
#include <vector>
using namespace std;

vector<int> mul(vector<int> &A,int b)
{
	vector<int> C;
	
	int t = 0;
	for(int i =0 ;i < A.size() || t;i++)
	{
		if(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]);
	
	 
}

高精度除法模板

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<int> div(vector<int> &A,int b,int &r)
{
	vector<int> C;
	r=0;
	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;	 
}

2. 前缀和
   

一维前缀和

#include <iostream>
using namespace std;

const int N = 100010;

int a[N],s[N];

int main()
{
	int n,m;
	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);
		printf("%d",s[r] - s[l-1]);
		
	}
	return 0;
	
}

二维前缀和

#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<=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]); // 算子矩阵的和 
	}
	return 0;
 } 

3. 差分

一维差分
差分是前缀和的逆

#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]);
	
	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]);
	
	return 0; 
}

二维差分

#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\n",b[i][j]);
		puts("");
	}
	return 0;
	
}