D. Time to Run

https://codeforces.com/contest/1301/problem/D

题意：给出n*m(1<=n,m<=500)格子，要求走k(1e9)步，问能否实现，能则输出步骤（<3000）.规则：同一个格子同一方向只能走一步.

解法：根据欧拉图可知，最多可走4*n*m-2*n-2*m步。

要使步骤尽可能少，就要是每一步尽可能多走，比如多走直线。

#include<bits/stdc++.h>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define cin(x) scanf("%lld" , &x);
using namespace std;
const int maxn = 1e3+9;
vector<pair<char , int> >ma;
vector<pair<char , int> >v;
void insert(char c , int m){
    if(m > 0) ma.pb(mp(c , m));
}

void solve(){
    int n , m , k ;
    cin >> n >> m >> k;
    if(4*n*m-2*n-2*m < k){
        cout << "NO" << endl;
        return ;
    }
    cout << "YES" << endl;
    insert('R' , m-1);
    rep(i , 1 , m-1){
        insert('D' , n-1);
        insert('U' , n-1);
        insert('L' , 1);
    }
    insert('D' , n-1);
    rep(i , 1 , n-1){
        insert('R' , m-1);
        insert('L' , m-1);
        insert('U' , 1);
    }
    int ans = 0 ;
    for(auto it : ma){
        if(k <= 0){
            break;
        }else{
            ans++;
            v.pb(mp(it.fi , k > it.se ? it.se : k));
            k -= it.se ;
        }
    }
    cout << ans << endl;
    for(auto it : v){
        cout << it.se << " " << it.fi << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    //int t ;
    //cin(t);
    //while(t--){
        solve();
    //}
}

4nm−