Submission #322173

#	Time	Username	Problem	Language	Result	Execution time	Memory
322173		zaneyu	Paint By Numbers (IOI16_paint)	C++14	32 / 100	1 ms	492 KiB

paint

/*input
........
2
3 4
*/
#include "paint.h"
#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<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define 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(int 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()
const ll maxn=1e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=(1e9+7);
const ll MOD2=1000002173;
const ld PI=acos(-1);
const ld eps=1e-9;
#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)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b,ll MOD){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
string solve_puzzle(string s, vector<int> c) {
    int n=sz(s);
    string cur="";
    REP(i,sz(c)){
        REP(j,c[i]) cur.pb('X');
        if(i!=sz(c)-1) cur.pb('_');
    }
    int left=n-sz(cur);
    string ans;
    ans.resize(n);
    REP(i,n){
        set<int> s;
        for(int j=i;j>=(i-left);j--){
            if(j<0 or j>=sz(cur)) s.insert('_');
            else s.insert(cur[j]);
        }
        if(sz(s)==2) ans[i]='?';
        else ans[i]=(*s.begin());
    }
    return ans;
}

• Subtask 1: 7 / 7
• Subtask 2: 3 / 3
• Subtask 3: 22 / 22
• Subtask 4: 0 / 27
• Subtask 5: 0 / 21
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10