제출 #698463

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
698463		sunwukong123	선물상자 (IOI15_boxes)	C++14	20 / 100	1 ms	340 KiB

boxes

#include <bits/stdc++.h>
using namespace std;
#define int long long 
#define FOR(i,s,e) for(int i = s; i <= (int)e; ++i)
#define DEC(i,s,e) for(int i = s; i >= (int)e; --i)
#define IAMSPEED ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef LOCAL
#define db(x) cerr << #x << "=" << x << "\n"
#define db2(x, y) cerr << #x << "=" << x << " , " << #y << "=" << y << "\n"
#define db3(a,b,c) cerr<<#a<<"="<<a<<","<<#b<<"="<<b<<","<<#c<<"="<<c<<"\n"
#define dbv(v) cerr << #v << ":"; for (auto ite : v) cerr << ite << ' '; cerr <<"\n"
#define dbvp(v) cerr << #v << ":"; for (auto ite : v) cerr << "{"  << ite.f << ',' << ite.s << "} "; cerr << "\n"
#define dba(a,ss,ee) cerr << #a << ":"; FOR(ite,ss,ee) cerr << a[ite] << ' '; cerr << "\n"
#define reach cerr << "LINE: " << __LINE__ << "\n";
#else
#define reach 
#define db(x)
#define db2(x,y)
#define db3(a,b,c)
#define dbv(v)
#define dbvp(v)
#define dba(a,ss,ee)
#endif
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define f first 
#define s second
#define g0(x) get<0>(x)
#define g1(x) get<1>(x)
#define g2(x) get<2>(x)
#define g3(x) get<3>(x)
typedef pair <int, int> pi;
typedef tuple<int,int,int> ti3;
typedef tuple<int,int,int,int> ti4;
int rand(int a, int b) { return a + rng() % (b-a+1); }
const int MOD = 1e9 + 7;
const int inf = (int)1e9 + 500;
const long long oo = (long long)1e18 + 500;
template <typename T> bool chmax(T& a, const T b) { return a<b ? a = b, 1 : 0; }
template <typename T> bool chmin(T& a, const T b) { return a>b ? a = b, 1 : 0; }
const int MAXN = 10000005;
int ans=oo;
int pref[MAXN]; // cost for the first i things to all be routed like 0->x->0
int suf[MAXN]; // cost for the last i things to all be routed like 0->y->0
int A[MAXN];
long long delivery(int32_t n, int32_t k, int32_t l, int32_t p[]) {
    reverse(p,p+n);
    int idx=n-1;
    while(idx >=0 && p[idx]==0){
        --idx;
    }
    if(idx == -1)return 0;

    reverse(p,p+idx+1);
    n=idx+1;
    

    FOR(i,1,n) {
        A[i] = p[i-1];
    }
    chmin(k,n);
    dba(A,1,n);
    db(l); 
    FOR(i,1,n) {
        pref[i] = pref[i-1];
        if(i % k == 0)pref[i] += 2*A[i];
    }
    int len = 0;
    DEC(i,n,1) {
        suf[i] = suf[i+1];
        ++len;
        if(len % k == 0)suf[i]+= 2 *(l-A[i]);
    }
    
    FOR(st,1,n-k+1) { // position of the start of the range i am looping
        int en=st+k-1; // ending position of the range that i am looping. 
        
        int fi=st-1;
        int sum = 0;
        sum += pref[fi];
        
        if(fi % k != 0)sum += 2*A[fi];

        int se=en+1;
        sum += suf[se];
        
       
        int len=(n-se+1);
        if(len % k != 0)sum+=2*(l-A[se]);

        sum += l;
        
        chmin(ans, sum);
    }
   
    FOR(i,0,n) {
        int sum = 0;
        sum += pref[i];
       
        if(i%k!=0)sum+=2*A[i];

        sum += suf[i+1];
        int len=(n-(i+1)+1);
        if(i!=n) {
            if(len % k != 0)sum+=2*(l-A[i+1]);
        }
        
        chmin(ans, sum);
    }

    return ans;
}
/*
int32_t main() {
    int n,m,k;cin>>n>>m>>k;
    int p[n];
    FOR(i,0,n-1)cin>>p[i];
    cout<<delivery(n,m,k,p);
}
*/

컴파일 시 표준 에러 (stderr) 메시지

boxes.cpp: In function 'long long int delivery(int32_t, int32_t, int32_t, int32_t*)':
boxes.cpp:57:10: warning: conversion from 'long long int' to 'int32_t' {aka 'int'} may change value [-Wconversion]
   57 |     n=idx+1;
      |       ~~~^~
boxes.cpp:90:13: warning: declaration of 'len' shadows a previous local [-Wshadow]
   90 |         int len=(n-se+1);
      |             ^~~
boxes.cpp:70:9: note: shadowed declaration is here
   70 |     int len = 0;
      |         ^~~
boxes.cpp:105:13: warning: declaration of 'len' shadows a previous local [-Wshadow]
  105 |         int len=(n-(i+1)+1);
      |             ^~~
boxes.cpp:70:9: note: shadowed declaration is here
   70 |     int len = 0;
      |         ^~~

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 0 / 15
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30