四、高精度：
1.大整数的存储
2.模拟加法的存储
123+89=212（Ai+Bi+t)
#include <vector>将数组的长度变长
例题
1.高精度减法
#include <iostream>
#include <vector>
using namespace std;
bool cmp(vector<int> &A, vector<int> &B)//判断是否有A>=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)// C = A - B, 满足A >= B, A >= 0, B >= 0
{
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();//去掉前面的0
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');
vector<int> C;
if (cmp(A, B)) C = sub(A, B);
else C = sub(B, A), cout << '-';
for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
cout << endl;
return 0;
}
2.高精度加法
#include <iostream>
#include <vector>
using namespace std;
const int base = 1000000000;
vector<int> add(vector<int> &A, vector<int> &B)
{
if (A.size() < B.size()) return add(B, A);
vector<int> C;
int t = 0;
for (int i = 0; i < A.size(); i ++ )
{
t += A[i];
if (i < B.size()) t += B[i];
C.push_back(t % base);
t /= base;
}
if (t) C.push_back(t);
return C;
}
int main()
{
string a, b;
vector<int> A, B;
cin >> a >> b;
for (int i = a.size() - 1, s = 0, j = 0, t = 1; i >= 0; i -- )
{
s += (a[i] - '0') * t;
j ++, t *= 10;
if (j == 9 || i == 0)
{
A.push_back(s);
s = j = 0;
t = 1;
}
}
for (int i = b.size() - 1, s = 0, j = 0, t = 1; i >= 0; i -- )
{
s += (b[i] - '0') * t;
j ++, t *= 10;
if (j == 9 || i == 0)
{
B.push_back(s);
s = j = 0;
t = 1;
}
}
auto C = add(A, B);
cout << C.back();
for (int i = C.size() - 2; i >= 0; i -- ) printf("%09d", C[i]);
cout << endl;
return 0;
}
五、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]);
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;
}
六、双指针算法
七、区间和与合并
例题
1.区间和
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
const int N = 300010;
int n, m;
int a[N], s[N];
vector<int> alls;//存入下标容器
vector<PII> add, query;//add增加容器，存入对应下标和增加的值的大小
//query存入需要计算下标区间和的容器
int find(int x)
{
int l = 0, r = alls.size() - 1;
while (l < r)//查找大于等于x的最小的值的下标
{
int mid = l + r >> 1;
if (alls[mid] >= x) r = mid;
else l = mid + 1;
}
return r + 1;//因为使用前缀和，其下标要+1可以不考虑边界问题
}
int main()
{
cin >> n >> m;
for (int i = 0; i < n; i ++ )
{
int x, c;
cin >> x >> c;
add.push_back({x, c});//存入下标即对应的数值c
alls.push_back(x);//存入数组下标x=add.first
}
for (int i = 0; i < m; i ++ )
{
int l, r;
cin >> l >> r;
query.push_back({l, r});//存入要求的区间
alls.push_back(l);//存入区间左右下标
alls.push_back(r);
}
// 区间去重
sort(alls.begin(), alls.end());
alls.erase(unique(alls.begin(), alls.end()), alls.end());
// 处理插入
for (auto item : add)
{
int x = find(item.first);//将add容器的add.secend值存入数组a[]当中，
a[x] += item.second;//在去重之后的下标集合alls内寻找对应的下标并添加数值
}
// 预处理前缀和
for (int i = 1; i <= alls.size(); i ++ ) s[i] = s[i - 1] + a[i];
// 处理询问
for (auto item : query)
{
int l = find(item.first), r = find(item.second);//在下标容器中查找对应的左右两端[l~r]下标，然后通过下标得到前缀和相减再得到区间a[l~r]的和
cout << s[r] - s[l - 1] << endl;
}
return 0;
}
2.区间和并
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
void merge(vector<PII> &segs)
{
vector<PII> res;
sort(segs.begin(), segs.end());
int st = -2e9, ed = -2e9;
for (auto seg : segs)
if (ed < seg.first)
{
if (st != -2e9) res.push_back({st, ed});
st = seg.first, ed = seg.second;
}
else ed = max(ed, seg.second);
if (st != -2e9) res.push_back({st, ed});
segs = res;
}
int main()
{
int n;
scanf("%d", &n);
vector<PII> segs;
for (int i = 0; i < n; i ++ )
{
int l, r;
scanf("%d%d", &l, &r);
segs.push_back({l, r});
}
merge(segs);
cout << segs.size() << endl;
return 0;
}