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; }
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); }
原文：https://www.cnblogs.com/uid001/p/11815825.html
