785. 快速排序
#include <iostream>
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) / 2], 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;
cin >> n;
for (int i = 0; i < n; ++i) cin >> q[i];
quick_sort(q, 0, n - 1);
for (int i = 0; i < n; ++i) cout << q[i] << " ";
puts("");
return 0;
}
786. 第k个数
#include <iostream>
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) << endl;
}
787. 归并排序
#include <iostream>
using namespace std;
const int N = 1e6 + 10;
int q[N], tmp[N], 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]) tmp[k++] = q[i++];
else tmp[k++] = q[j++];
}
while (i <= mid) tmp[k++] = q[i++];
while (j <= r) tmp[k++] = q[j++];
for (int i = l, j = 0; i <= r; i++, j++) q[i] = tmp[j];
}
int main () {
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;
}
788. 逆序对的数量
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
int a[N], tmp[N];
LL merge_sort(int q[], int l, int r)
{
if (l >= r) return 0;
int mid = l + r >> 1;
LL 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
{
res += mid - i + 1;
tmp[k ++ ] = q[j ++ ];
}
while (i <= mid) tmp[k ++ ] = q[i ++ ];
while (j <= r) tmp[k ++ ] = q[j ++ ];
for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];
return res;
}
int main()
{
int n;
scanf("%d", &n);
for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
cout << merge_sort(a, 0, n - 1) << endl;
return 0;
}
789. 数的范围
#include <iostream>
using namespace std;
const int N = 100010;
int q[N], m, n;
int main () {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i) scanf("%d", &q[i]);
while (m--) {
int x;
scanf("%d", &x);
int l = 0, r = n - 1;
while (l < r) {
int mid = l + r >> 1;
if (q[mid] >= x) r = mid;
else l = mid + 1;
}
if (q[l] != x) {
cout << "-1 -1" << endl;
} else {
cout << l << " ";
int l = 0, r = n - 1;
while (l < r) {
int mid = l + r + 1 >> 1;
if (q[mid] <= x) l = mid;
else r = mid - 1;
}
cout << l << endl;
}
}
}
790. 数的三次方根
`#include <iostream>
using namespace std;
double n;
int main() {
cin >> n;
double l = -10000, r = 10000;
while (r - l >= 1e-7) {
double mid = (l + r) / 2;
if (mid * mid * mid >= n) r = mid;
else l = mid;
}
printf("%.6lf\n", l);
}
795. 前缀和
#include <iostream>
using namespace std;
const int N = 100010;
int n, m, q[N], s[N];
int main() {
cin >> n >> m;
for (int i = 1; i <= n; ++i) {
cin >> q[i];
s[i] = s[i - 1] + q[i];
}
int a, b;
for (int i = 0; i < m; ++i) {
cin >> a >> b;
cout << s[b] - s[a - 1] << endl;
}
return 0;
}
- 子矩阵的和
#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q, s[N][N];
int main() {
cin >> n >> m >> q;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) cin >> s[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];
}
while (q--) {
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
cout << s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1] << endl;
}
}
799. 最长连续不重复子序列
#include <iostream>
using namespace std;
int n;
const int N = 100010;
int a[N], s[N];
int main() {
cin >> n;
for (int i = 0; i < n; ++i) cin >> a[i];
int res = 0;
for (int i = 0, j = 0; i < n; ++i) {
s[a[i]]++;
while (s[a[i]] > 1) {
s[a[j]]--;
j++;
}
res = max(res, i - j + 1);
}
cout << res << endl;
return 0;
}
800. 数组元素的目标和
#include <iostream>
using namespace std;
const int N = 100010;
int n, m, x, a[N], b[N];
int main() {
cin >> n >> m >> x;
for (int i = 0; i < n; ++i) cin >> a[i];
for (int i = 0; i < m; ++i) cin >> b[i];
for (int i = 0, j = m - 1; i < n; ++i) {
while (j >= 0 && a[i] + b[j] > x) j--;
if (j >= 0 && a[i] + b[j] == x) cout << i << " " << j << endl;
}
return 0;
}
803. 区间合并
#include <iostream>
#include <algorithm>
#include <vector>
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;
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() {
cin >> n;
for (int i = 0; i < n; ++i) {
int a, b;
cin >> a >> b;
segs.push_back({a, b});
}
merge(segs);
cout << segs.size() << endl;
return 0;
}
