Submission #441474

#	Time	Username	Problem	Language	Result	Execution time	Memory
441474		zaneyu	Remittance (JOI19_remittance)	C++14	100 / 100	477 ms	8588 KiB

remittance

/*input
2
4 1
0 0
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("Ofast,unroll-loops")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<int,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll INF64=1e18+1;
const int INF=0x3f3f3f3f;
int MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-3;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
inline ll mult(ll a,ll b){
    if(a>=MOD) a%=MOD;
    if(b>=MOD) b%=MOD;
    return (a*b)%MOD;
}
inline ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
const int maxn=1e6+5;
const int maxlg=__lg(maxn)+2;
int a[maxn],b[maxn];
int32_t main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n;
    cin>>n;
    REP(i,n) cin>>a[i]>>b[i];
    int p=0;
    REP(i,n){
        if(a[i]-b[i]>a[p]-b[p]) p=i;
    }
    REP(asd,30){
        for(int i=p;i!=(p-1+n)%n;i=(i+1)%n){
            int nx=(i+1)%n;
            if(a[i]<b[i]){
                continue;
            }
            int z=(a[i]-b[i]+1)/2;
            if((a[i]-b[i])%2 and b[i]==0) z--;
            a[i]-=2*z;
            a[nx]+=z;
        }
        int i=(p-1+n)%n;
                    int nx=(i+1)%n;
            if(a[i]<b[i]){
                continue;
            }
            int z=(a[i]-b[i]+1)/2;
            if((a[i]-b[i])%2 and b[i]==0) z--;
            a[i]-=2*z;
            a[nx]+=z;
    }
    REP(i,n) if(a[i]!=b[i]){
        cout<<"No";
        return 0;
    }
    cout<<"Yes";
}  

• Subtask 1: 15 / 15
• Subtask 2: 40 / 40
• Subtask 3: 45 / 45