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