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Codeforces Round #264 (Div. 2)

题目链接

A：注意特判正好的情况，其他就一个个去判断记录最大值即可

B：扫一遍，不够的用钱去填即可，把多余能量记录下来

C：把主副对角线处理出来，然后黑格白格只能各选一个最大的放即可

D：转化为DAG最长路问题，每个数字记录下在每个序列的位置，如果一个数字能放上去，那么肯定是每个序列上的数字都是在之前最末尾数字的后面

E：大力出奇迹，预处理出树，然后每次查询从当前位置一直往上找到一个符合的即可，如果没有符合的就是-1

代码：

A：

#include #include #include using namespace std;int n, s;int main() {	scanf("%d%d", &n, &s);	int x, y;	int flag = 1;	int ans = 0;	for (int i = 0; i < n; i++) {		scanf("%d%d", &x, &y);		if (x == s) {			if (y == 0) {				ans = max(ans, y);				flag = 0;			}		}		else if (x < s) {			ans = max(ans, (100 - y) % 100);			flag = 0;		}	}	if (flag) printf("-1\n");	else printf("%d\n", ans);	return 0;}

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B：

#include #include const int N = 100005;typedef long long ll;int n;ll h[N];int main() {	scanf("%d", &n);	for (int i = 1; i <= n; i++)		scanf("%lld", &h[i]);	ll now = 0;	ll ans = 0;	for (int i = 1; i <= n; i++) {		if (h[i] > h[i - 1]) {			ll need = h[i] - h[i - 1];			if (now >= need) {				now -= need;			} else {				ans += need - now;				now = 0;			}		} else {			now += h[i - 1] - h[i];		}	}	printf("%lld\n", ans);	return 0;}

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C：

#include #include const int N = 2005;typedef long long ll;int n;ll g[N][N], zhu[N + N], fu[N + N];int main() {	scanf("%d", &n);	for (int i = 0; i < n; i++)		for (int j = 0; j < n; j++)			scanf("%lld", &g[i][j]);	for (int i = 0; i < n; i++) {		for (int j = 0; j < n; j++) {			zhu[i - j + n] += g[i][j];			fu[i + j] += g[i][j];		}	}	ll b = -1, w = -1;	int bx, by, wx, wy;	for (int i = 0; i < n; i++) {		for (int j = 0; j < n; j++) {			ll sum = zhu[i - j + n] + fu[i + j] - g[i][j];			if ((i + j) % 2 == 0) {				if (b < sum) {					b = sum;					bx = i + 1;					by = j + 1;				}			}			else {				if (w < sum) {					w = sum;					wx = i + 1;					wy = j + 1;				}			}		}	}	printf("%lld\n", b + w);	printf("%d %d %d %d\n", bx, by, wx, wy);	return 0;}

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D：

#include #include #include using namespace std;const int N = 1005;struct Num {	int v[10], la;} num[N];bool cmp(Num a, Num b) {	return a.la < b.la;}int n, k, dp[N][N];bool judge(int i, int j) {	for (int x = 0; x < k; x++) {		if (num[i].v[x] < num[j].v[x]) return false;	}	return true;}int main() {	scanf("%d%d", &n, &k);	for (int i = 0; i < k; i++) {		int tmp;		for (int j = 1; j <= n; j++) {			scanf("%d", &tmp);			num[tmp].v[i] = j;		}	}	for (int i = 0; i <= n; i++) {		int maxv = 0;		for (int j = 0; j < k; j++) {			maxv = max(maxv, num[i].v[j]);		}		num[i].la = maxv;	}	sort(num, num + n + 1, cmp);	/*	for (int i = 0; i <= n; i++) {		for (int j = 0; j < k; j++) {			printf("%d ", num[i].v[j]);		}		printf("\n");	}*/	for (int i = 1; i <= n; i++) {		for (int j = 0; j < i; j++) {			dp[i][j] = max(dp[i][j], dp[i - 1][j]);			if (judge(i, j)) {				dp[i][i] = max(dp[i][i], dp[i][j] + 1);			}		}	}	int ans = 0;	for (int i = 0; i <= n; i++)		ans = max(ans, dp[n][i]);	printf("%d\n", ans);	return 0;}

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E：

#include #include #include using namespace std;const int N = 100005;int n, q, val[N], f[N];vector g[N];void dfs(int u, int fa) {	f[u] = fa;	for (int i = 0; i < g[u].size(); i++) {		int v = g[u][i];		if (v == fa) continue;		dfs(v, u);	}}int gcd(int a, int b) {	while (b) {		int tmp = a % b;		a = b;		b = tmp;	}	return a;}int query(int u) {	int v = val[u];	u = f[u];	while (u) {		if (gcd(val[u], v) > 1) return u;		u = f[u];	}	return -1;}int main() {	scanf("%d%d", &n, &q);	for (int i = 1; i <= n; i++)		scanf("%d", &val[i]);	int u, v;	for (int i = 1; i < n; i++) {		scanf("%d%d", &u, &v);		g[u].push_back(v);		g[v].push_back(u);	}	dfs(1, 0);	int tp, a, b;	while (q--) {		scanf("%d%d", &tp, &a);		if (tp == 1) {			printf("%d\n", query(a));		} else {			scanf("%d", &b);			val[a] = b;		}	}	return 0;}

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