arc079

D. Decrease (Contestant ver.)

大意: 每次操作选一个最大数$-n$,其余数全$+1$. 要求构造一个序列$a$, 使得恰好$k$次操作后最大值不超过$n-1$.

只要让$k$次操作以后恰好变全为$n-1$即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#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, 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


ll a[60];

int main() {
    ll k;
    cin>>k;
    ll p = k/50, r = k-50*p;
    int n = 50;
    REP(i,1,n) a[i] = p+49;
    REP(i,1,r) a[i]+=n-r+1;
    REP(i,r+1,n) a[i]-=r;
    printf("%d\n",n);
    REP(i,1,n) printf("%lld ",a[i]);hr;
}

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E. Decrease (Judge ver.)

大意: 给定序列$a$, 求进行多少次$D$题中的操作后, 最大值不超过$n-1$.

暴力模拟

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#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, 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



#ifdef __APPLE__
const int N = 1e2+50;
#else
const int N = 1e6+50;
#endif


int n;
ll a[N];

int calc() {
    int ans = 0;
    while (1) {
        sort(a+1,a+1+n);
        if (a[n]<=n-1) break;
        ++ans;
        a[n]-=n;
        REP(i,1,n-1) ++a[i];
    }
    return ans;
}

int main() {
    scanf("%d", &n);
    REP(i,1,n) scanf("%lld",a+i);
    ll ans = 0;
    while (1) {
        sort(a+1,a+1+n);
        if (a[1]>60) {
            ll t = a[1]-60;
            ans += t*n;
            REP(i,1,n) a[i]-=t;
        }
        if (a[n]<=2000) { 
            ans += calc();
            break;
        }
        ll ma = -1;
        int p = 0;
        REP(i,2,n) if (a[i]-a[i-1]>ma) ma = a[i]-a[i-1], p = i;
        if (ma<=200) throw;
        ll w = n-p+1, t = (a[p]-a[p-1])/(n+1);
        ans += t*w;
        REP(i,1,p-1) a[i] += t*w;
        REP(i,p,n) a[i] += t*(-n+w-1);
    }
    printf("%lld\n", ans);
}

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