文章目录
- 快速排序
- 归并排序
- 前缀和
- 差分
- 双指针算法
快速排序
快排
基本思想
using namespace std;
const int N = 1000010;
int q[N];
void quick_sort(int q[], int l, int r){
if(l >= r) return;
int x = q[l + r >> 1], i = l - 1, j = r + 1;
while(i < j){
do i++; while(q[i] < x);
do j--; while(q[j] > x);
if(i < j) swap(q[i], q[j]);
}
quick_sort(q, l, j);
quick_sort(q, j+1, r);
}
int main(){
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d", &q[i]);
quick_sort(q, 0, n - 1);
for(int i = 0; i < n; i++) printf("%d ", q[i]);
return 0;
}
求第k小的数
using namespace std;
const int N = 100010;
int n, k, q[N];
int quick_sort(int l, int r, int k){
if(l >= r) return q[l];
int i = l - 1, j = r + 1, x = q[l + r >> 1];
while(i < j){
do i++; while(q[i] < x);
do j--; while(q[j] > x);
if(i < j) swap(q[i], q[j]);
}
int sl = j - l + 1;
if(k <= sl) return quick_sort(l, j, k);
return quick_sort(j + 1, r, k - sl);
}
int main(){
scanf("%d%d", &n, &k);
for(int i = 0; i < n; i++) scanf("%d", &q[i]);
cout << quick_sort(0, n - 1, k);
return 0;
}
归并排序
归并排序
using namespace std;
const int N = 1e5 + 10;
int q[N], temp[N];
void merge_sort(int q[], int l, int r){
if(l >= r) return;
int mid = (l + r) >> 1;
merge_sort(q, l, mid), merge_sort(q, mid + 1, r);
int k = 0, i = l, j = mid + 1;
while(i <= mid && j <= r){
if(q[i] <= q[j]) temp[k++] = q[i++];
else temp[k++] = q[j++];
}
while(i <= mid) temp[k++] = q[i++];
while(j <= r) temp[k++] = q[j++];
for(i = l,k = 0; i <= r; i++, k++) q[i] = temp[k];
}
int main(){
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d", &q[i]);
merge_sort(q, 0, n - 1);
for(int i = 0; i < n; i++) printf("%d ", q[i]);
return 0;
}
求逆序对的个数
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
int q[N], tmp[N];
LL merge_sort(int q[], int l, int r){
if(l >= r) return 0;
int mid = l + r >> 1;
LL res = 0;
res += (merge_sort(q, l, mid) + merge_sort(q, mid + 1, r));
int k = 0, i = l, j = mid + 1;
while(i <= mid && j <= r){
if(q[i] <= q[j]) tmp[k++] = q[i++];
else{
tmp[k++] = q[j++];
res += (mid - i + 1); // 此时前半段剩下的数都严格大于当前q[j]
}
}
while(i <= mid) tmp[k++] = q[i++];
while(j <= r) tmp[k++] = q[j++];
for(i = l, k = 0; i <= r; i++, k++) q[i] = tmp[k];
return res;
}
int main(){
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d", &q[i]);
cout << merge_sort(q, 0, n-1) << endl;
return 0;
}
整数二分
根据分界点是红色这边还是绿色这边分为两个板子
1是红色,2是绿色
板子1
int l = 0, r = n - 1;
while(l < r){
int mid = l + r + 1 >> 1;
if(check(mid)) l = mid;
else r = mid - 1;
}
板子2
int l = 0, r = n - 1;
while (l < r) {
int mid = l + r >> 1;
if (check(mid)) l = mid + 1;
else r = mid;
}
浮点数二分
精度为4位小数的时候 < 1e-6,5位的时候 < 1e-7,6位时,,相当于精度比题目要求多 1e-2 就基本不会出问题!
第一种利用
第二种直接 循环迭代 次
拿求 的三次方跟举例:
double l = -100, r = 100;
while(r - l > 1e-8){
double mid = (l + r) / 2;
if(mid * mid * mid < x) l = mid;
else r = mid;
}
高精度
高精度加法
using namespace std;
vector<int> add(vector<int> &A, vector<int> &B){
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 != 0) C.push_back(t);
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;
}
高精度加法压位(压9位)
using namespace std;
const int base = 1000000000;
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 % 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, j = 0, t = 1, s = 0; i >= 0; i--) {
s += (a[i] - '0') * t;
j++;
t *= 10;
if(i == 0 || j == 9){
A.push_back(s);
j = 0;
s = 0;
t = 1;
}
}
for(int i = b.size() - 1, j = 0, t = 1, s = 0; i >= 0; i--) {
s += (b[i] - '0') * t;
j++;
t *= 10;
if(i == 0 || j == 9){
B.push_back(s);
j = 0;
s = 0;
t = 1;
}
}
auto C = add(A, B);
printf("%d", C.back());
for(int i = C.size() - 2; i >= 0; i--) printf("%09d", C[i]);
return 0;
}
高精度减法
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;
int t = 0;
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; // 1 或者 -1 (取决于上面是+t还是-t)
else t = 0;
}
while(C.size() > 1 && C.back() == 0) C.pop_back(); // 去掉前导0
return C;
}
int main(){
string a, b; // a = "12345"
vector<int> A, B; // A = [5, 4, 3, 2, 1]
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{
cout << "-";
C = sub(B, A);
}
for(int i = C.size() - 1; i >= 0; i--) cout << C[i];
cout << endl;
return 0;
}
高精度乘法
using namespace std;
vector<int> mul(vector<int> &A, int b){
vector<int> C;
int t = 0;
for(int i = 0; i < A.size(); i++){
t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
while(t > 0) C.push_back(t % 10), t /= 10; // 可以与上面的for循环合并
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');
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
using namespace std;
// A / b = C 余数为 r
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()); // 数组反转
while(C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main(){
string a; // '12345'
int b;
cin >> a >> b;
vector<int> A; // [5, 4, 3, 2, 1]
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--) cout << C[i];
cout << endl << r;
return 0;
}
前缀和
下标从1开始
一维前缀和
using namespace std;
const int N = 100010;
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];
while(m--){
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", s[r] - s[l - 1]);
}
return 0;
}
二维前缀和(子矩阵的和)
注意: 需要把每个格子当成一个坐标点!!!
using namespace std;
const int N = 1010;
int a[N][N], s[N][N];
int main(){
int n, m, q;
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];
while(q--){
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int z = s[x2][y2] - s[x1-1][y2] - s[x2][y1-1] + s[x1-1][y1-1];
printf("%d\n", z);
}
return 0;
}
差分
一维差分
using namespace std;
const int N = 100010;
int a[N], b[N];
void insert(int l, int r, int c){
b[l] += c;
b[r+1] -= c;
}
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++) insert(i, i, a[i]); // 构造差分数组
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(差分矩阵)
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[x1][y2+1] -= c;
b[x2+1][y1] -= 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;
scanf("%d%d%d%d%d", &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 ", b[i][j]);
}
puts("");
}
return 0;
}
双指针算法
- 先写朴素做法
- 根据双指针模版+本题的性质将 “1.” 改为双指针算法(看和之间有没有单调关系)
最长连续不重复子序列
using namespace std;
const int N = 100010;
int a[N], s[N]; // s记录每个数出现的次数
int main(){
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d", &a[i]);
int res = 0;
for(int i = 0, j = 0; i < n; i++){
s[a[i]]++;
while(j <= i && s[a[i]] > 1){ // 本题这里 j <= i可以不写
s[a[j]]--;
j++;
}
res = max(res, i - j + 1);
}
cout << res << endl;
return 0;
}
数组元素的目标和
using namespace std;
const int N = 1e5 + 10;
int a[N], b[N];
int main(){
int n, m, x;
scanf("%d%d%d", &n, &m, &x);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 1; i <= m; i++) scanf("%d", &b[i]);
for(int i = 1, j = m; i <= n; i++){
while(j && a[i] + b[j] > x) j--;
if(a[i] + b[j] == x){
printf("%d %d", (i-1), (j-1));
break;
}
}
return 0;
}
位运算
的二进制表示中第 k 位是几?
n >> k & 1
求一个十进制数的二进制可以采用循环加上述操作计算出来
操作:返回 的最后一位
(~)
在 C++ 中 表示为补码也就是 ~
using namespace std;
int main(){
int n;
cin >> n;
int x, res = 0;
while(n--){
cin >> x;
res = 0;
while(x > 0) x -= (x & -x), res ++;
cout << res << ' ';
}
return 0;
}
离散化
using namespace std;
typedef pair<int, int> PII;
const int N = 300010;
int a[N], s[N];
// 记录离散化后的操作(排序+去重)
// 某操作在alls的所在位置的下标+1 即是某操作映射到原数组的位置
vector<int> alls;
// 记录所有的添加和查询操作
vector<PII> add, query;
int find(int x){
int l = 0, r = alls.size() - 1;
// 二分找出第一个大于等于x的值
while(l < r){
int mid = l + r >> 1;
if(alls[mid] <= x) r = mid;
else l = mid + 1;
}
// +1 是因为 a[] 下标从1开始。
return r + 1;
}
int main(){
int n, m;
scanf("%d%d", &n, &m);
int x, y;
for(int i = 0; i < n; i++){
scanf("%d%d", &x, &y);
add.push_back({x, y});
alls.push_back(x);
}
for(int i = 0; i < m; i++){
scanf("%d%d", &x, &y);
query.push_back({x, y});
alls.push_back(x);
alls.push_back(y);
}
// 对操作进行排序 + 去重
sort(alls.begin(), alls.end());
alls.erase(unique(alls.begin(), alls.end()), alls.end());
// 遍历添加操作, 找到离散话后的位置,进行操作
for(int i = 0; i < n; i++){
x = add[i].first;
y = add[i].second;
a[find(x)] += y;
}
// 预处理前缀和
for(int i = 1; i <= alls.size(); i++){
s[i] = s[i-1] + a[i];
}
for(int i = 0; i < m; i++){
x = query[i].first;
y = query[i].second;
int l = find(x);
int r = find(y);
cout << s[r] - s[l - 1] << endl;
}
return 0;
}
区间合并
using namespace std;
typedef pair<int, int> PII;
const int N = 100010;
int n;
vector<PII> segs;
void merge(vector<PII> &segs){
vector<PII> res;
// pair<a, b> 默认先以a排序,a相同时再以b排序
sort(segs.begin(), segs.end());
int st = -2e9, ed = -2e9;
for(PII seg: segs){
if(seg.first > ed){
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(){
cin >> n;
for(int i = 0; i < n; i++){
int l, r;
cin >> l >> r;
segs.push_back({l, r});
}
merge(segs);
cout << segs.size() << endl;
return 0;
}