C. Coloring Trees
O(n^4)暴力DP就好了
#include <iostream>
#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 '
'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7;
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;}
//head
const ll INF = 0x3f3f3f3f3f3f3f3f;
int n, m, k;
ll dp[111][111][111];
int c[111][111], a[111];
void chkmin(ll &x, ll y) {x=min(x,y);}
int main() {
scanf("%d%d%d", &n, &m, &k);
if (m==1&&k>1) return puts("-1");
REP(i,1,n) scanf("%d", a+i);
REP(i,1,n) REP(j,1,m) scanf("%d", &c[i][j]);
memset(dp,0x3f,sizeof dp);
dp[0][1][1] = 0;
REP(i,1,n) REP(j,1,k) REP(pre,1,m) {
int nxt;
if (a[i]) {
if (i==1) nxt=1;
else nxt=j+(pre!=a[i]);
chkmin(dp[i][nxt][a[i]],dp[i-1][j][pre]);
continue;
}
REP(now,1,m) {
if (i==1) nxt=1;
else nxt=j+(pre!=now);
chkmin(dp[i][nxt][now],dp[i-1][j][pre]+c[i][now]);
}
}
ll ans = INF;
REP(j,1,m) chkmin(ans,dp[n][k][j]);
printf("%lld
", ans==INF?-1:ans);
}
D. Directed Roads
大意: n个点的有向图, 每个点出度为1, 每条边的方向可以改变, 问有多少种方案使得图无环.
每个点出度为1, 那么图是一个基环树森林. 再观察一下可以发现每个连通块的答案为$2^n-2^{n-环上点数+1}$
#include <iostream>
#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 '
'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, 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;}
//head
const int N = 1e6+10;
int n, ans, sz, tot;
int dfn[N], fa[N];
vector<int> g[N];
void dfs(int x) {
dfn[x] = ++*dfn,++tot;
for (int y:g[x]) {
if (dfn[y]) {
if (dfn[y]<dfn[x]) continue;
for (; y!=x; y=fa[y]) ++sz;
}
else fa[y]=x, dfs(y);
}
}
int main() {
scanf("%d", &n);
REP(i,1,n) {
int t;
scanf("%d", &t);
g[t].pb(i);
g[i].pb(t);
}
int ans = 1;
REP(i,1,n) if (!dfn[i]) {
tot=sz=0,dfs(i);
ll t = qpow(2,tot)-qpow(2,tot-sz);
ans = ans*t%P;
}
if (ans<0) ans+=P;
printf("%d
", ans);
}
E. ZS and The Birthday Paradox
大意: 假设一年有$2^n$天，问$k$个小朋友中有两个小朋友生日相同的概率
显然答案为$1-frac{2^n}{2^n}frac{2^n-1}{2^n}cdotsfrac{2^n-k+1}{2^n}$
因为模数比较小, 分子可以暴力求出. gcd显然为2的幂, 统计一下再除去gcd即可
#include <iostream>
#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 '
'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e6+3, 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;}
//head
ll n, k;
int main() {
scanf("%lld%lld", &n, &k);
if (n<=61&&(1ll<<n)<k) return puts("1 1"),0;
ll cnt = n%(P-1);
for (ll i=(k-1)>>1; i; i>>=1) (cnt+=i)%=(P-1);
cnt = P-1-cnt;
cnt = qpow(2,cnt);
ll x = qpow(2,n), A=1, B = qpow(x,k);
if (k-1>=x) A = 0;
else REP(i,0,k-1) A=A*(x-i)%P;
A = A*cnt%P, B = B*cnt%P;
A = (B-A)%P;
if (A<0) A+=P;
printf("%lld %lld
",A,B);
}