牛客网暑期ACM多校训练营（第五场）
牛客网暑期ACM多校训练营（第五场）
A. gpa
二分答案，然后就转化为是否满足
\(\frac {\sum s[i]c[i]}{\sum s[i]} ≥ D\),
\(\sum s[i]c[i] ≥ \sum s[i]D\),
\(\sum s[i](c[i]-D) ≥ 0\)
显然科目越少gpa越高，于是去掉最小的k个判断即可。
#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define PII pair<int,int>
#define MP make_pair
typedef long long ll;
const int N = 1e5 + 7;
using namespace std;
int n,k;
struct node{
int s,c;
node(){}node(int a,int b) {c=a;s=b;}
}a[N];
// 1. 2^n 暴力的做法
int ct(int s) {
int ans = 0;
for(int i=s;i;i-=i&-i)++ans;
return ans;
}
double cal_1() {
double ans = 0;
rep(s,0,(1<<n)-1) if(ct(s)>=n-k) {
int t1 = 0, t2 = 0;
rep(i,0,n-1) if(s&(1<<i)) {
t1 += a[i+1].s*a[i+1].c;
t2 += a[i+1].s;
}
ans = max(ans,(1.0*t1)/t2);
}
return ans;
}
//**
double c[N];
double solve(double x) {
double ans = 0;
rep(i,1,n) c[i] = a[i].s*(a[i].c-x);
sort(c+1,c+1+n);
rep(i,k+1,n) ans+=c[i];
return ans;
}
double cal_2() {
double l = 0, r = 1001.0,mid;
rep(ti,1,30) {
mid = (l+r)*0.5;
if(solve(mid)>=0) l=mid;
else r=mid;
}
return l;
}
int main() {
srand(time(0));
scanf("%d%d",&n,&k);
rep(i,1,n) scanf("%d",&a[i].s);
rep(i,1,n) scanf("%d",&a[i].c);
// printf("%.10f\n",cal_1());
printf("%.10f\n",cal_2());
return 0;
}
E. room
考虑新的第i组人住在原来的第j个房间，不改变的人就是两个集合的交。然后二分图最大权匹配即可。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
using namespace std;
const int N = 305;
const int inf = 0x3f3f3f3f;
int nx,ny,n;
int g[N][N],linker[N],lx[N],ly[N];
int slack[N];
bool visx[N], visy[N];
bool dfs(int x) {
visx[x] = 1;
for(int y = 0; y < ny; ++y) {
if(visy[y])continue;
int tmp = lx[x] + ly[y] - g[x][y];
if(tmp == 0) {
visy[y] = 1;
if(linker[y] == -1||dfs(linker[y])) {
linker[y] = x;
return 1;
}
}
else if(slack[y] > tmp)
slack[y] = tmp;
}
return 0;
}
int KM() {
memset(linker,-1,sizeof(linker));
memset(ly,0,sizeof(ly));
for(int i = 0; i < nx; ++i) {
lx[i] = -inf;
for(int j = 0; j < ny; ++j)
if(g[i][j]>lx[i])
lx[i]=g[i][j];
}
for(int x = 0; x < nx; ++x) {
for(int i = 0; i < ny; ++i)
slack[i] = inf;
while(1) {
memset(visx,0,sizeof(visx));
memset(visy,0,sizeof(visy));
if(dfs(x)) break;
int d = inf;
for(int i = 0; i < ny; ++i)
if(!visy[i] && d > slack[i])
d = slack[i];
for(int i = 0; i < nx; ++i)
if(visx[i])
lx[i] -= d;
for(int i = 0; i < ny; ++i) {
if(visy[i]) ly[i] += d;
else slack[i] -= d;
}
}
}
int res = 0;
for(int i = 0; i < ny; ++i)
if(linker[i] != -1)
res += g[linker[i]][i];
return res;
}
int a[301][4],b[301][4];
map<int,int> M;
int cal(int a[],int b[]) {
M.clear(); int ans = 0;
for (int i = 0; i < 4; ++i) M[a[i]]=1;
for (int i = 0; i < 4; ++i) ans += M[b[i]];
return ans;
}
int main() {
while (~scanf("%d", &n)) {
for (int i = 0; i < n; ++i)
for (int j = 0; j < 4; ++j)
scanf("%d",&a[i][j]);
for (int i = 0; i < n; ++i)
for (int j = 0; j < 4; ++j)
scanf("%d",&b[i][j]);
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
g[i][j] = cal(a[i],b[j]);
nx = ny = n;
printf("%d\n", 4*n-KM());
}
return 0;
}
F.take
考虑每个答案的贡献，就是\(p_i \Pi_{j<i,a_j>a_i} (1-pj)\)，前缀乘积用树状数组维护。
#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define pb push_back
typedef long long ll;
const int N = 2e5 + 7;
const ll mod = 998244353;
using namespace std;
ll q_pow(ll a,ll b) {
ll ans = 1;
while(b) {
if(b&1) ans = (ans*a)%mod;
a=(a*a)%mod;
b >>= 1LL;
}
return ans;
}
ll inv;
vector<ll> v;
int getid(ll x) {
return lower_bound(v.begin(),v.end(),x) - v.begin() + 1;
}
int cc;
ll B[N],b[N];
ll n;
ll p[N],a[N];
void init() {
rep(i,0,2e5+1)B[i]=1;
}
ll ask(int x) {
ll ans = 1;
for(int i=x;i;i-=i&-i)ans=(ans*B[i])%mod;
return ans;
}
void update(int x,ll v) {
for(int i=x;i<=cc;i+=i&-i) B[i]=(B[i]*v)%mod;
}
int main() {
inv = q_pow(100LL,mod-2);
scanf("%lld",&n);
rep(i,1,n) scanf("%lld%lld",&p[i],&a[i]),v.pb(a[i]),p[i]=(p[i]*inv)%mod;
sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end());
cc = v.size();
init();
ll ans = 0;
rep(i,1,n) {
ans = (ans + (p[i]*ask(100000-getid(a[i])))%mod)%mod;
update(100001-getid(a[i]),(1-p[i]+mod)%mod);
}
printf("%lld\n",ans);
return 0;
}
G. max
相当于在[1,n/c]，找最大的两个互质的数，相邻的两个数一定互质，再特判一下1即可。
#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
typedef long long ll;
using namespace std;
ll c,n;
int main() {
scanf("%lld%lld",&c,&n);
ll a = (n/c)*c;
ll b = (n/c-(n/c!=1))*c;
if(__gcd(a,b)==c&&a>=1&&a<=n&&b>=1&&b<=n) printf("%lld\n",a*b);
else puts("-1");
return 0;
}
J. plan
尽量先选性价比最高的，特判几种情况就行了
#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
typedef unsigned long long ll;
using namespace std;
ll n, p2, p3, ans = 0;
int main() {
scanf("%lld %lld %lld", &n, &p2, &p3);
if(2*p3 >= 3*p2) {
if(n%2 == 0) {
ans = n/2*p2;
}
else {
ans = (n/2+1)*p2;
ans = min(ans,(n/2)*p2+p3);
if(n>=3)ans = min(ans,(n-3)/2*p2+p3);
}
}
else {
if(n%3==0)
ans = n/3*p3;
else if(n%3==1) {
ans = ((n-1)/3+1)*p3;
ans = min(ans,((n-1)/3)*p3+p2);
if(n>=4) ans = min(ans,(n-4)/3*p3+2*p2);
}
else if(n%3==2) {
ans = (n/3+1)*p3;
ans = min(ans,(n/3)*p3+p2);
if(n>=2) ans = min(ans,(n-2)/3*p3+p2);
}
}
printf("%lld\n",ans);
return 0;
}
