ACwing算法基础课_2
原文链接:here
高精度加法
// C = A + B, A >= 0, B >= 0
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;
}
高精度减法
//判断是否有A>=B
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, 满足A >= B, A >= 0, B >= 0
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;
}
高精度乘低精度
// C = A * b, A >= 0, b > 0
vector<int> mul(vector<int> &A, int b) {
vector<int> C;
int t = 0;
for (int i = 0; i < A.size() || t; i ++ ) {
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
高精度除以低精度
// A / b = C ... r, A >= 0, b > 0
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;
}
一维前缀和
S[i] = a[1] + a[2] + ... a[i]
a[l] + ... + a[r] = S[r] - S[l - 1]
二维前缀和
s[i][j] = 第i行j列格子左上部分所有元素的和
二维前缀和:
s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+a[i][j]
同理,对于任意一个边长为R的正方形,我们有:
a[i][j]=s[i][j]-s[i-R][j]-s[i][j-R]+s[i-R][j-R]
以(x1, y1)为左上角,(x2, y2)为右下角的子矩阵的和为:
s[x2][y2]-s[x1 -1][y2]-s[x2][y1 - 1]+s[x1-1][y1-1]
一维差分
对于一个给定的数列A,他的差分数列B定义为:B[1]=A[1],B[i]=A[i]-A[i-1](2<=i<=n)
=>A[i]=A[i-1]+B[i]
给区间[l, r]中的每个数加上c:B[l] += c, B[r + 1] -= c
二维差分
不难=>A[i][j]=A[i-1][j]+A[i][j-1]-A[i-1][j-1]+S[i][j](类比一维差分,此处S相当于一维差分的B)
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