Codeforces Round #333 (Div. 2)

水 A - Two Bases

水题，但是pow的精度不高，应该是转换成long long精度丢失了干脆直接double就可以了。被hack掉了。用long long能存的下

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int b1[44], b2[44];
int n, m;
 
int main(void)  {
    scanf ("%d%d", &n, &m);
    double ans1 = 0, ans2 = 0;
    for (int i=1; i<=n; ++i) {
        scanf ("%d", &b1[i]);
    }
    for (int i=n; i>=1; --i) {
        ans1 += pow ((double) m, n-i) * b1[i];
    }
    scanf ("%d%d", &n, &m);
    for (int i=1; i<=n; ++i) {
        scanf ("%d", &b2[i]);
    }
    for (int i=n; i>=1; --i) {
        ans2 += pow ((double) m, n-i) * b2[i];
    }
    if (ans1 < ans2) puts ("<");
    else if (ans1 > ans2)    puts (">");
    else    puts ("=");
 
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int b1[44], b2[44];
int n, m;

ll _pow(int m, int x)	{
	ll ret = 1;
	for (int i=1; i<=x; ++i)	{
		ret *= m;
	}
	return ret;
}

int main(void)	{
	cout << (ll) pow (39, 9) << endl;
	cout << _pow (39, 9) << endl;
	scanf ("%d%d", &n, &m);
	ll ans1 = 0, ans2 = 0;
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &b1[i]);
	}
	for (int i=n; i>=1; --i)	{
		ans1 += _pow (m, n-i) * b1[i];
	}
	scanf ("%d%d", &n, &m);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &b2[i]);
	}
	for (int i=n; i>=1; --i)	{
		ans2 += _pow (m, n-i) * b2[i];
	}
	if (ans1 < ans2)	puts ("<");
	else if (ans1 > ans2)	puts (">");
	else	puts ("=");

	return 0;
}

/*
9 39
10 20 16 36 30 29 28 9 8
9 38
12 36 10 22 6 3 19 12 34
*/

尺取法 B - Approximating a Constant Range

简单说就是维护[i, j]区间的最大值和最小值以及它们的最后的位置。

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int a[N];

int main(void)	{
	int n;	scanf ("%d", &n);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &a[i]);
	}
	int ans = 1, mn = a[1], mx = a[1], i = 1, j = 2;
	int p1 = 1, p2 = 1;
	while (j <= n)	{
		if (a[j] <= mn)	{
			mn = a[j];	p1 = j;
		}
		else if (a[j] >= mx)	{
			mx = a[j];	p2 = j;
		}
		
		if (mx - mn <= 1)	{
			ans = max (ans, j - i + 1);
		}
		else	{
			while (mx - mn > 1)	{
				if (p1 < p2 && p1 + 1 <= j)	{
					mn = a[p1+1];	i = p1 + 1;	p1++;
				}
				else if (p1 >= p2 && p2 + 1 <= j)	{
					mx = a[p2+1];	i = p2 + 1;	p2++;
				}
				else	break;
			}
			ans = max (ans, j - i + 1);
		}
		j++;
	}
	
	printf ("%d
", ans);

	return 0;
}

BFS C - The Two Routes

两点之间要不是地铁要不就是汽车，那么1到n也一样，只要一次BFS就行了，水水的。

我刚做了一道双向BFS，想套一个试试，写麻烦了，但是还是A掉了，想想还是有问题，vis数组有问题，数据水了。。因为有一个一定会在一步到达,和另一个不冲突

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 4e2 + 10;
const int M = N * N;
const int INF = 0x3f3f3f3f;
struct Edge	{
	int v, w, nex;
	Edge() {}
	Edge(int v, int w, int nex) : v (v), w (w), nex (nex) {}
}edge[M];
int head[N];
bool lk[N][N];
bool vis[N][N][2];
bool vis2[N][2];
bool ok[2];
int n, m, e;
queue<int> que[2];

void init(void)	{
	memset (head, -1, sizeof (head));
	e = 0;
}

void add_edge(int u, int v, int w)	{
	edge[e].v = v;	edge[e].w = w;	edge[e].nex = head[u];
	head[u] = e++;
}

bool BFS(int typ, int tim)	{
	int sz = que[typ].size ();
	while (sz--)	{
		int u = que[typ].front ();	que[typ].pop ();
		for (int i=head[u]; ~i; i=edge[i].nex)	{
			int v = edge[i].v, w = edge[i].w;
			if (w != typ)	continue;
			if (v == n)	{
				ok[typ] = true;
				if (ok[typ^1])	return true;
			}
            if (vis2[v][typ])    continue;
            vis2[v][typ] = true;
			if (vis[v][tim][typ^1])	continue;
			vis[v][tim][typ] = true;
			que[typ].push (v);
		}
	}
	return false;
}

int run(void)	{
	que[0].push (1);	que[1].push (1);
	int step = 0;
	while (!que[0].empty () || !que[1].empty ())	{
		step++;
		if (step > 800)	break;
        if (!ok[0]) {
            if (BFS (0, step))	return step;    
        }
		if (!ok[1]) {
            if (BFS (1, step))	return step;
        }
	}
	return -1;
}

int main(void)	{
	init ();
	scanf ("%d%d", &n, &m);
	for (int u, v, i=1; i<=m; ++i)	{
		scanf ("%d%d", &u, &v);
		add_edge (u, v, 1);
		add_edge (v, u, 1);
		lk[u][v] = lk[v][u] = true;
	}
	int cnt = 0;
	for (int i=1; i<=n; ++i)	{
		for (int j=i+1; j<=n; ++j)	{
			if (i == j || lk[i][j])	continue;
			cnt++;
			add_edge (i, j, 0);
			add_edge (j, i, 0);
		}
	}
	if (cnt == 0)	{
		puts ("-1");	return 0;
	}
	printf ("%d
", run ());

	return 0;
}

找规律+区间端点 D - Lipshitz Sequence

题意：q次询问，询问[l, r]的所有子区间 sum (max (abs (a[i] - a[j]) / j - i)) (1 <= i < j <= n)

分析：首先要知道最大值是abs (a[i] - a[i-1])，然后要弄出多少个子区间包含了这个值，用到KMP思想，left[i]/right[i]能记录i最左边和最右边不大于该值的位置，左闭右开防止重复

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int a[N];
int lef[N], righ[N];
int n, m;

ll run(int l, int r)	{
	if (l == r)	return 0;
	for (int i=l+1; i<=r; ++i)	{
		int j = i - 1;
		while (j > l && abs (a[j] - a[j-1]) <= abs (a[i] - a[i-1]))	j = lef[j];
		lef[i] = j;
	}
	for (int i=r; i>l; --i)	{
		int j = i + 1;
		while (j <= r && abs (a[j] - a[j-1]) < abs (a[i] - a[i-1]))	j = righ[j];
		righ[i] = j;
	}
	ll ret = 0;
	for (int i=l+1; i<=r; ++i)	{
		ret += 1ll * (i - lef[i]) * (righ[i] - 1 - i + 1) * abs (a[i] - a[i-1]);
	}
	return ret;
}

int main(void)	{
	scanf ("%d%d", &n, &m);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &a[i]);
	}
	for (int l, r, i=1; i<=m; ++i)	{
		scanf ("%d%d", &l, &r);
		printf ("%I64d
", run (l, r));
	}
	

	return 0;
}

概率DP E - Kleofáš and the n-thlon

题意：n次比赛，m个人，现在给出n次比赛一个人的排名，问他最后的排名的期望

分析：转换成其余人得分少于他的概率 * 总人数 +１，dp[i][j] 表示前i次比赛，得分为j时的期望。dp[i][j] = sum + dp[i-1][j-1] - dp[i-1][j-m-1] - dp[i-1][j-a[i]]；状态转移用前缀和优化复杂度

#include <bits/stdc++.h>
using namespace std;

double dp[105][105*1005];
int a[105];

int main(void)	{
	int n, m;	scanf ("%d%d", &n, &m);
	int sum = 0;
	for (int i=0; i<n; ++i)	{
		scanf ("%d", &a[i]);	sum += a[i];
	}
	if (m == 1)	{
		printf ("%.10f
", 1.0);	return 0;
	}
	dp[0][0] = m-1;
    for (int i=1; i<=n; ++i) {
        double sum = 0;
        for (int j=i; j<=i*m; ++j) {
            sum += dp[i-1][j-1] / (m-1);
            if (j - 1 - m >= 0) sum -= dp[i-1][j-m-1] / (m-1);
            dp[i][j] = sum;
            if (j - a[i-1] >= 0) dp[i][j] -= dp[i-1][j-a[i-1]] / (m-1);
        }
    }
	double ans = 1;
    for (int i = 0; i < sum; i++) ans += dp[n][i];
    printf("%.10f
", ans);

	return 0;
}