最近状态差到爆炸.
(AT)连掉两把分，啥时候能上黄啊。
(A)
考虑直接动归。
把(O(n^2))的动归后缀和优化成(O(n))
A
#include<iostream>
#include<cstdio>
#define ll long long
#define N 100005
#define mod 1000000007
ll a[N],f[N],g[N],sf[N],sg[N];
ll n,ans,sum;
int main(){
scanf("%lld",&n);
for(int i = 1;i <= n;++i)
scanf("%lld",&a[i]),sum = (sum + a[i]) % mod;
//f:i为-时从i到n有多少种合法方案，g:这些合法方案的权数。
f[n] = 1,g[n] = (-2 * a[n] + mod) % mod;
f[n + 1] = 1;
sf[n] = 2,sf[n + 1] = 1,sg[n] = g[n];
for(int i = n - 1;i >= 2;--i){
f[i] = sf[i + 2];
g[i] = (f[i] * (-2 * a[i] + mod) % mod + sg[i + 2]) % mod;
sf[i] = (sf[i + 1] + f[i]) % mod;
sg[i] = (sg[i + 1] + g[i]) % mod;
}
// for(int i = n;i >= 2;--i)
// std::cout<<f[i]<<" "<<g[i]<<std::endl;
for(int i = n;i >= 2;--i)
ans = (ans + sum * f[i] % mod + g[i]) % mod;
ans = (ans + sum) % mod;
std::cout<<ans<<std::endl;
}
(B)
听说B是一个结论题，正解来看呢，应该是把(n)个可取值都试一遍，但是我写的模拟退火。
B
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<cmath>
#define ll long long
#define N 100005
struct P{int x,y,w;}p[N];
ll n;
double ansx,ansy;
inline double find(double x){
double ans = 0;
for(int i = 1;i <= n;++i)
ans += (p[i].x + x - std::min((double)p[i].x,(double)(2 * x))) / (1.0 * n);
return ans;
}
int main(){
// freopen("q.in","r",stdin);
// freopen("q.out","w",stdout);
srand(273352);
scanf("%lld",&n);
for(int i = 1;i <= n;++i){
scanf("%d",&p[i].x);
ansx += p[i].x; }
ansx = (ansx) / (1.0 * n);
ansy = 0x7f7f7f7f;
double eps = 1e-15;
double T = 200;//初始温度
while(T > eps && ((double)(clock())/CLOCKS_PER_SEC)<1.9){//终止态
// std::cout<<T<<" "<<(rand() * 2 - RAND_MAX) * T<<std::endl;
double nowx = ansx + ((rand() * 2 - RAND_MAX + 1) * T);//在值域[ansx - t,ansx + t];中挑选一个随机数
long double z = find(nowx) - find(ansx);
if(z < 0)
ansx = nowx,ansy = std::min(ansy,find(nowx));
else
if(exp(-z / T) * RAND_MAX > rand())//随机接受
ansx = nowx;
T *= 0.997;//降温速率
// std::cout<<ansx<<std::endl;
}
printf("%.12lf
",ansy);
}
(C)
考虑每一个数在(fib)数系下都有唯一分解。
做完了。
C
#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
ll f[200],to[200],ti[200];
ll ans[200];
ll n;
ll k;
//0 1
//1 1
//1 2
//1 3
//4 3
//4 7
int main(){
scanf("%lld",&n);
ll q = n;
ll len = 1;
f[0] = 1,f[1] = 1;
while(f[len] <= n + 1){
f[++len] = f[len - 1] + f[len - 2];
}
for(int i = len;i >= 1;--i)
if(n >= f[i])
to[++to[0]] = i,n -= f[i];
std::sort(to + 1,to + to[0] + 1);
for(int i = 1;i <= to[0];++i)
ti[i] = to[to[0]] - to[i];
std::sort(ti + 1,ti + to[0] + 1);
ll s = to[to[0]];
ll x = 0,y = 0;
ll now = 1;
for(int i = 0;i <= s;++i){
if(i == ti[now]){
if(i & 1)
ans[++k] = 1,x += 1;
else
ans[++k] = 2,y += 1;
++now;
}
if(i & 1)
ans[++k] = 4,y += x;
else
ans[++k] = 3,x += y;
}
std::cout<<k<<std::endl;
if(x == q){
for(int i = 1;i <= k;++i)
std::cout<<ans[i]<<std::endl;
}
else{
for(int i = 1;i <= k;++i){
if(ans[i] <= 2)
std::cout<<3 - ans[i]<<std::endl;
else
std::cout<<7 - ans[i]<<std::endl;
}
}
}
(D)
考虑如果后手已经想好的每个数对，那么其实游戏进程没有差别。
所以这是一个假博弈。
我们考虑对每一位进行操作，如果这一位的(0),(1)的数量都是偶数，那么递归进子树操作。
否则则把左子树和右子树左右匹配，用(01tire)找出最小的匹配,因为再往下递归的所有对都小于这个匹配。
D
#include<iostream>
#include<cstdio>
#define ll long long
ll to[12000005][2];
ll cnt[12000005];
ll n;
ll dfncnt;
inline void insert(ll x){
ll u = 0;
cnt[u] ++ ;
for(int i = 29;i >= 0;--i){
int t = (x >> i) & 1;
if(!to[u][t])
to[u][t] = ++ dfncnt;
u = to[u][t];
cnt[u] ++ ;
}
}
ll ans = 0,tmp;
#define l(u) to[u][0]
#define r(u) to[u][1]
inline void find(ll p1,ll p2,ll now,ll dep){
// std::cout<<p1<<" "<<p2<<" "<<now<<" "<<dep<<std::endl;
if(dep == 0){
tmp = std::min(now,tmp);
return;
}
bool q = 0;
for(int i = 0;i <= 1;++i)
if(to[p1][i] && to[p2][i]){
q = 1;
find(to[p1][i],to[p2][i],now,dep - 1);
}
if(q)
return;
for(int i = 0;i <= 1;++i)
if(to[p1][i] && to[p2][!i]){
find(to[p1][i],to[p2][!i],now | (1 << (dep - 1)),dep - 1);
q = true;
}
}
inline void dfs(int u,int dep){
// std::cout<<u<<" "<<dep<<std::endl;
if(!to[u][1] && !to[u][0])
return;
if(cnt[l(u)] % 2 && cnt[r(u)]){
tmp = (1 << 30);
find(l(u),r(u),(1 << (dep - 1)),dep - 1);
ans = std::max(ans,tmp);
return ;
}
if(l(u))
dfs(l(u),dep - 1);
if(r(u))
dfs(r(u),dep - 1);
}
int main(){
scanf("%lld",&n);
for(int i = 1;i <= 2 * n;++i){
ll x;
scanf("%lld",&x);
insert(x);
}
dfs(0,30);
std::cout<<ans<<std::endl;
}
(E)
由于(lcm(x,y) = frac{x * y}{gcd(x,y)})
考虑最后一个数，那么有(gcd(lcm(a_j),a_i) < a_i)
即(lcm（gcd(a_i,a_j)) < a_i)这里是由于精度所以不能选择前一种（
然后依次从后向前选择就好了。
E
#include<iostream>
#include<cstdio>
#define ll long long
#define N 305
ll ans[N],a[N];
bool vis[N];
ll n;
inline ll g(ll a,ll b){return (b == 0) ? a : g(b,a % b);}
int main(){
scanf("%lld",&n);
for(int i = 1;i <= n;++i)
scanf("%lld",&a[i]);
for(int i = n;i >= 1;--i){
bool q;
for(int j = 1;j <= n;++j){
ll lcm = 1;
ll gcd = 1;
if(!vis[j]){
q = 1;
for(int k = 1;k <= n;++k){
if(!vis[k] && k != j){
gcd = g(a[k],a[j]);
lcm = lcm / g(gcd,lcm) * gcd;
if(lcm >= a[j]){
q = 0;
break;
}
}
}
if(q){
ans[i] = a[j];
vis[j] = 1;
break;
}
}
}
if(!q){
puts("No");
return 0;
}
}
puts("Yes");
for(int i = 1;i <= n;++i)
std::cout<<ans[i]<<" ";
}