Wannafly挑战赛23

B. 游戏

大意: $n$堆石子, 第$i$堆初始$a_i$, 每次只能选一堆, 假设一堆个数$x$, 只能取$x$的约数, 求先手第一步必胜取法.

SG入门题, 预处理出所有$SG$值. 先手要必胜必须满足留给后手的异或值为0.

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

const int N = 1e5+10;

int n, a[N], sg[N], vis[N];

vector<int> fac[N];

void init() {

	REP(i,1,N-1) for(int j=i;j<N;j+=i) fac[j].pb(i);

	REP(i,1,N-1) {

		for (int j:fac[i]) vis[sg[i-j]]=i;

		REP(j,0,N-1) if (vis[j]!=i) {

			sg[i] = j; break;

		}

	}

}

int main() {

	init();

	scanf("%d", &n);

	int s = 0;

	REP(i,1,n) {

		scanf("%d", a+i);

		s ^= sg[a[i]];

	}

	if (!s) return puts("0"),0;

	int ans = 0;

	REP(i,1,n) {

		for (int x:fac[a[i]]) {

			if (!(s^sg[a[i]]^sg[a[i]-x])) ++ans;

		}

	}

	printf("%d\n", ans);

}

C.收益

大意: 要融资$L$元$n$个人, 融资成功收益$M$, 第$i$个客户有$p_i$概率出钱$m_i$元, 若融资成功要付$m_ir_i$元, 求最后期望收益.

DP求出最终得到$x$元的概率及期望花费, 然后统计答案.

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

const int N = 111;

int n, L, M;

int m[N], r[N], p[N];

int f[2][500100], g[2][500100];

int main() {

	scanf("%d%d%d", &n, &L, &M);

	int sum = 0, inv100 = inv(100);

	REP(i,1,n) {

		scanf("%d%d%d",m+i,r+i,p+i);

		sum += m[i];

		r[i] = (ll)m[i]*r[i]%P*inv100%P;

		p[i] = (ll)p[i]*inv100%P;

	}

	int cur = 0;

	f[cur][0] = 1;

	REP(i,1,n) {

		cur ^= 1;

		REP(j,0,sum) {

			int nxt = j-m[i]>=0?j-m[i]:sum+1;

			f[cur][j] = ((ll)f[!cur][nxt]*p[i]+(ll)f[!cur][j]*(1-p[i]))%P;

			g[cur][j] = (((ll)g[!cur][nxt]+(ll)r[i]*f[!cur][nxt])%P*p[i]+(ll)g[!cur][j]*(1-p[i]))%P;

		}

	}

	int ans = 0;

	REP(i,L,sum) ans = (ans+(ll)f[cur][i]*M-g[cur][i])%P;

	if (ans<0) ans += P;

	printf("%d\n", ans);

}

D.漂亮的公园

给定树, 点$i$颜色为$c[i]$, 每次询问所有颜色为$x$的点到颜色为$y$的点的最大距离.

结论: 对于树上点集$S$, $S$内距离最远的两点为$x,y$, 则其他点$u$到点集$S$的最远距离必然是$u$到$x$或$u$到$y$.

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <queue>

#include <map>

#define REP(i,a,b) for(int i=a;i<=b;++i)

using namespace std;

const int N = 1e5+10;

int n, q, c[N], sz[N], dep[N];

int fa[N], son[N], top[N];

int f[N][2];

vector<int> g[N];

int b[N];

void dfs(int x, int d, int f) {

    sz[x]=1,fa[x]=f,dep[x]=d;

    for (int y:g[x]) if (y!=f) {

        dfs(y,d+1,x),sz[x]+=sz[y];

        if (sz[y]>sz[son[x]]) son[x]=y;

    }

}

void dfs(int x, int tf) {

    top[x]=tf;

    if (son[x]) dfs(son[x],tf);

    for (int y:g[x]) if (!top[y]) dfs(y,y);

}

int lca(int x, int y) {

    while (top[x]!=top[y]) {

        if (dep[top[x]]<dep[top[y]]) swap(x,y);

        x=fa[top[x]];

    }

    return dep[x]<dep[y]?x:y;

}

int dis(int x, int y) {

	if (!x||!y) return 0;

	return dep[x]+dep[y]-2*dep[lca(x,y)];

}

void upd(int x) {

	int &A = f[c[x]][0], &B = f[c[x]][1];

	if (!A) A = x;

	else if (!B) B = x;

	else {

		int d1 = dis(A,B), d2 = dis(A,x), d3 = dis(B,x);

		if (d2>d1&&d2>d3) {

			if (d2>d3) B = x;

			else A = x;

		}

		else if (d3>d1) A = x;

	}

}

int main() {

	scanf("%d%d", &n, &q);

	REP(i,1,n) scanf("%d",c+i),b[i]=c[i];

	sort(b+1,b+1+n),*b=unique(b+1,b+1+n)-b-1;

	REP(i,1,n) c[i]=lower_bound(b+1,b+1+*b,c[i])-b;

	REP(i,2,n) {

		int u, v;

		scanf("%d%d", &u, &v);

		g[u].push_back(v);

		g[v].push_back(u);

	}

	dfs(1,0,0),dfs(1,1);

	REP(i,1,n) upd(i);

	while (q--) {

		int x, y;

		scanf("%d%d", &x, &y);

		int xx=lower_bound(b+1,b+1+*b,x)-b;

		int yy=lower_bound(b+1,b+1+*b,y)-b;

		if (b[xx]!=x||b[yy]!=y) {

			puts("0"); continue;

		}

		x = xx, y = yy;

		int ans = 0;

		REP(i,0,1) REP(j,0,1) ans = max(ans, dis(f[x][i],f[y][j]));

		printf("%d\n", ans);

	}

}

E. 排序

大意: 初始有一个长为$2n$的排列, 随机打乱后, 将奇数位升序排列, 求期望逆序对. $n\le 5e7$

打个表发现答案是$\frac{7n^2-n}{12}$, 官方题解如下

F 生成树计数留坑

巴特西

Wannafly挑战赛23

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