股票买卖II
动态规划解
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int f[N][2];
int w[N];
int n;
int main() {
cin >> n;
for (int i = 1; i <= n; i ++ ) cin >> w[i];
//第0天不能买进, 不然第一天就会被卖出;
f[0][1] = -0x3f3f3f3f;
for (int i = 1; i <= n; i ++ ) {
f[i][1] = max(f[i - 1][1], f[i - 1][0] - w[i]);
f[i][0] = max(f[i - 1][1] + w[i], f[i - 1][0]);
}
cout << max(f[n][0], f[n][1]) << endl;
return 0;
}
贪心解
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N];
int n;
int main() {
cin >> n;
for (int i = 1; i <= n; i ++ ) cin >> a[i];
int sum = 0;
for (int i = 1; i <= n; i ++ ) {
if (a[i] >= a[i + 1]) continue;
else sum += a[i + 1] - a[i];
}
cout << sum << endl;
return 0;
}
股票买卖III
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int f[N][3][2];
int n;
int w[N];
int main() {
cin >> n;
for (int i = 1; i <= n; i ++ ) cin >> w[i];
memset(f, -0x3f, sizeof f);
f[0][0][0] = 0;
for (int i = 1; i <= n; i ++ ) {
for (int j = 0; j <= 2; j ++ ) {
f[i][j][0] = f[i - 1][j][0];
//f[i][j][0] = max(f[i - 1][j][0], f[i - 1][j - 1][1] + w[i]);
//只有前面有过一次完整交易时,才可以递推过来
if (j)
f[i][j][0] = max(f[i][j][0], f[i - 1][j - 1][1] + w[i]);
f[i][j][1] = max(f[i - 1][j][1], f[i - 1][j][0] - w[i]);
}
}
int res = 0;
for (int j = 0; j <= 2; ++ j) res = max(res, f[n][j][0]);
cout << res << endl;
return 0;
}
股票买卖VII
解法1
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, M = 110;
int f[N][M][2];
int w[N];
int n, m;
int main() {
cin >> n >> m;
for (int i = 1; i <= n; i ++ ) cin >> w[i];
memset(f, -0x3f, sizeof f);
f[0][0][0] = 0;
for (int i = 1; i <= n; i ++ ) {
for (int j = 0; j <= m; j ++ ) {
f[i][j][0] = f[i - 1][j][0];
if (j) f[i][j][0] = max(f[i][j][0], f[i - 1][j - 1][1] + w[i]);
f[i][j][1] = max(f[i - 1][j][0] - w[i], f[i - 1][j][1]);
}
}
int res = 0;
for (int i = 0; i <= m; i ++ ) res = max(res, f[n][i][0]);
cout << res << endl;
return 0;
}
解法2
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, M = 110;
int f[N][M][2];
int w[N];
int n, m;
int main() {
cin >> n >> m;
for (int i = 1; i <= n; i ++ ) cin >> w[i];
memset(f, -0x3f, sizeof f);
for (int i = 0; i <= n; i ++ )
f[i][0][0] = 0;
for (int i = 1; i <= n; i ++ ) {
for (int j = 1; j <= m; j ++ ) {
f[i][j][0] = max(f[i - 1][j][0], f[i - 1][j][1] + w[i]);
f[i][j][1] = max(f[i - 1][j - 1][0] - w[i], f[i - 1][j][1]);
}
}
int res = 0;
for (int i = 0; i <= m; i ++ ) res = max(res, f[n][i][0]);
cout << res << endl;
return 0;
}