标签：ace target pop fine 表示 over fas 枚举 lin
https://codeforces.com/contest/1244/problem/C
求出满足以下公式的
给出n,p,w,dn,p,w,d，求解：
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(ll i=(a);i<=(b);++i)
#define dep(i,a,b) for(ll i=(a);i>=(b);--i)
#define pb push_back
typedef long long ll;
const int maxn=(int)2e5+100;
const int mod=(int)1e9+7;
ll n,p,d,w;
int main(){
scanf("%lld%lld%lld%lld",&n,&p,&w,&d);
if(n*w<p) return puts("-1"),0;
rep(i,0,w){
if((p-i*d)%w==0){
ll x=(p-1ll*i*d)/w,y=i;
if(x+y>n) return puts("-1"),0;
return printf("%lld %lld %lld",x,y,n-x-y),0;
}
if(i*d>p) break;
}
return puts("-1"),0;
}
https://codeforces.com/contest/1244/problem/D
给出三行数据，第一行表示选择颜色1的各个顶点的花费，第二行表示颜色2...
然后每三个相邻点颜色不能一样
容易发现只有链的情况合法，链的情况就很简单了。
因为颜色个数较少，所以直接枚举颜色排列染色就行。
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define dep(i,a,b) for(int i=(a);i>=(b);--i)
#define pb push_back
typedef long long ll;
const int maxn=(int)2e5+100;
const int mod=(int)1e9+7;
int n,ind[maxn],pos=1,pp=1,id[10][maxn];
ll c[4][maxn];
vector<int> g[maxn<<1];
ll dfs(int pos,int st,int fa,int col1,int col2){
ll ans=0;int col;
rep(i,1,3) if(i!=col1&&i!=col2) col=i;
id[pos][st]=col;
if((int)g[st].size()==1) return c[col-1][st];
rep(i,0,1) if(g[st][i]!=fa) return c[col-1][st]+dfs(pos,g[st][i],st,col2,col);
}
int main(){
scanf("%d",&n);
rep(i,1,n) scanf("%lld",&c[0][i]);
rep(i,1,n) scanf("%lld",&c[1][i]);
rep(i,1,n) scanf("%lld",&c[2][i]);
rep(i,1,n-1){
int u,v;scanf("%d%d",&u,&v);
g[u].pb(v);g[v].pb(u);
ind[u]++;ind[v]++;
}
int num=0,st=0,ed=0;
rep(i,1,n){
if(ind[i]==1){
num++;
if(st) ed=i;
else st=i;
}
else if(ind[i]>2) return puts("-1"),0;
}
if(num!=2) return puts("-1"),0;
ll ans=(ll)1e18;
int st2=g[st][0],st3;
rep(i,0,1) if(g[st2][i]!=st) st3=g[st2][i];
rep(i,1,3) rep(j,1,3) if(i!=j){
pos++;
id[pos][st]=i;id[pos][st2]=j;
ll tmp=c[i-1][st]+c[j-1][st2]+dfs(pos,st3,st2,i,j);
if(ans>tmp) ans=tmp,pp=pos;
}
printf("%lld\n",ans);
rep(i,1,n) printf("%d ",id[pp][i]);
puts("");
}
https://codeforces.com/contest/1244/problem/E
#include <bits/stdc++.h>
#define MP make_pair
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
// #define Local
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e5 + 5;
ll n, k;
int a[N];
ll sum[N];
bool chk(int d) {
for(int i = 1; i <= n; i++) {
int low = lower_bound(a + 1, a + n + 1, a[i]) - a - 1;
int high = upper_bound(a + 1, a + n + 1, a[i] + d) - a;
ll tot = 1ll * low * a[i] - sum[low] + sum[n] - sum[high - 1] - 1ll * (n - high + 1) * (a[i] + d);
if(tot <= k) return true;
}
for(int i = 1; i <= n; i++) {
int high = upper_bound(a + 1, a + n + 1, a[i]) - a;
int low = lower_bound(a + 1, a + n + 1, a[i] - d) - a - 1;
ll tot = 1ll * low * (a[i] - d) - sum[low] + sum[n] - sum[high - 1] - 1ll * (n - high + 1) * a[i];
if(tot <= k) return true;
}
return false;
}
void run() {
int res=0;
for(int i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
int l = 0, r = 1e9 + 1, mid;
while(l <=r) {
mid = (l + r) >> 1;
if(chk(mid)) res=mid,r=mid-1;
else l= mid+1;
}
cout << res<< ‘\n‘;
}
int main() {
while(cin >> n >> k) run();
return 0;
}
https://codeforces.com/contest/1244/problem/G
sum=∑i=max(pi,qi)≤k，sum要最大
最小的sum肯定是（n）*（n+1）/2；
考虑将第一个排列顺序排放。
然后考虑第二个排列刚开始也是顺序排放的。
然后考虑交换位置视为元素左移和元素右移。
显然，元素右移不会对答案产生影响。
但是元素左移，左移几个位置它就产生多少贡献。
那么显然可以发现这个变化是连续的，直接贪心即可。
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << ‘ ‘; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ‘ ‘; err(args...); }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << ‘ ‘; pt(args...); }
template <template<typename...> class T, typename t, typename... A>
void pt(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ‘ ‘; pt(args...); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
//head
constexpr int N = 1e6 + 10;
int n, a[N]; ll k;
void run() {
ll base = 1ll * n * (n + 1) / 2;
if (base > k) return pt(-1);
for (int i = 1; i <= n; ++i) a[i] = i;
ll remind = k - 1ll * n * (n + 1) / 2;
int low = 1;
for (int i = n; i >= 1 && i > low; --i) {
if (i - low <= remind) {
remind -= i - low;
swap(a[i], a[low]);
++low;
} else {
swap(a[i], a[i - remind]);
remind = 0;
break;
}
}
pt(k - remind);
for (int i = 1; i <= n; ++i) cout << i << " \n"[i == n];
for (int i = 1; i <= n; ++i) cout << a[i] << " \n"[i == n];
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
while (cin >> n >> k) run();
return 0;
}
Codeforces Round #592 (Div. 2)
标签：ace target pop fine 表示 over fas 枚举 lin
原文地址：https://www.cnblogs.com/hgangang/p/11919758.html
