【算法2-1】前缀和与差分

P2671 [NOIP2015 普及组] 求和

题目链接：P2671 [NOIP2015 普及组] 求和 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include 
#define mod 10007
using namespace std;
int num[100010], color[100010], s[100010][2], sum[100010][2];

int main() {
	int n, m, ans = 0;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> num[i];
	}
	for (int i = 1; i <= n; i++) {
		cin >> color[i];
		s[color[i]][i % 2]++;
		sum[color[i]][i % 2] = (sum[color[i]][i % 2] + i) % mod;
	}
	for (int i = 1; i <= n; i++) {
		ans = (ans + num[i] * (i * (s[color[i]][i % 2] - 2) % mod + sum[color[i]][i % 2]) % mod) % mod;
	}
	cout << ans;
	return 0;
}

P1115 最大子段和

题目链接：P1115 最大子段和 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include 
#include 
using namespace std;
int a[200010], sum[200010];

int main() {
	int n, ans = -(1 << 29);
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		if (i == 0) {
			sum[i] = a[i];
		} else {
			sum[i] = max(a[i], sum[i - 1] + a[i]);
		}
		ans = max(ans, sum[i]);
	}
	cout << ans;
	return 0;
}

P3397 地毯

题目链接：P3397 地毯 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include 
using namespace std;
int a[1005][1005];

int main() {
	int n, m, x1, y1, x2, y2;
	cin >> n >> m;
	while (m--) {
		cin >> x1 >> y1 >> x2 >> y2;
		for (int i = x1; i <= x2; i++) {
			for (int j = y1; j <= y2; j++) {
				a[i][j]++;
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cout << a[i][j] << ' ';
		}
		cout << endl;
	}
	cout << endl;
}

P3406 海底高铁

题目链接：P3406 海底高铁 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include 
#include 
using namespace std;
long long p[100010], c[100010][3], a[100010], f[100010];

int main() {
	unsigned long long n, m, ans = 0;
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		cin >> p[i];
	}
	for (int i = 1; i <= n - 1; i++) {
		cin >> c[i][0] >> c[i][1] >> c[i][2];
	}
	for (int i = 0; i < m - 1; i++) {
		if (p[i] < p[i + 1]) {
			a[p[i]]++;
			a[p[i + 1]]--;
		} else {
			a[p[i + 1]]++;
			a[p[i]]--;
		}
	}
	for (int i = 1; i <= n - 1; i++) {
		if (i == 1) {
			f[i] = a[i];
		} else {
			f[i] = f[i - 1] + a[i];
		}
		ans += min(f[i] * c[i][0], c[i][2] + f[i] * c[i][1]);
	}
	cout << ans;
	return 0;
}

P1719 最大加权矩形

题目链接：P1719 最大加权矩形 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include 
#include 
#include 
using namespace std;
int a[150][150], t[150], dp[150];

int main() {
	int n, ans = -(1 << 29);
	cin >> n;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> a[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		memset(t, 0, sizeof(t));
		for (int j = i; j <= n; j++) {
			for (int k = 1; k <= n; k++) {
				t[k] += a[j][k];
			}
			memset(dp, 0, sizeof(dp));
			for (int r = 1; r <= n; r++) {
				dp[r] = max(t[r], dp[r - 1] + t[r]);
				ans = max(ans, dp[r]);
			}
		}
	}
	cout << ans;
	return 0;
}

P2004 领地选择

题目链接：P2004 领地选择 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include 
#include 
using namespace std;
int a[1010][1010], f[1010][1010];

int main() {
	int n, m, c, x, y, maxx = -(1 << 29);
	cin >> n >> m >> c;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			cin >> a[i][j];
			f[i][j] = a[i][j] + f[i][j - 1] + f[i - 1][j] - f[i - 1][j - 1];
		}
	}
	for (int i = c; i <= n; i++) {
		for (int j = c; j <= m; j++) {
			if (f[i][j] - f[i][j - c] - f[i - c][j] + f[i - c][j - c] > maxx) {
				maxx = f[i][j] - f[i][j - c] - f[i - c][j] + f[i - c][j - c];
				x = i - c + 1;
				y = j - c + 1;
			}
		}
	}
	cout << x << ' ' << y;
	return 0;
}

P5638 【CSGRound2】光骓者的荣耀

题目链接：P5638 【CSGRound2】光骓者的荣耀 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include 
#include 
using namespace std;
long long a[1000010], f[1000010];

int main() {
	long long n, k, maxx = 0;
	cin >> n >> k;
	for (int i = 1; i <= n - 1; i++) {
		cin >> a[i];
		f[i] = f[i - 1] + a[i];
	}
	for (int i = 0; i + k <= n - 1; i++) {
		if (f[i + k] - f[i] > maxx) {
			maxx = f[i + k] - f[i];
		}
	}
	cout << f[n - 1] - maxx;
	return 0;
}