[SWEA] 3074 입국심사

Time Lapse :30min 50sec

3074.cpp

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#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;

#define FIR(x) fastreadINT_byReference(x)
#define FIV() fastreadINT_byValue()
#define gc() getchar_unlocked()

inline void fastreadINT_byReference(int &ret){
int N = gc(), flag = 1;
for(;N<'0'||N>'9';N=gc())
if(N=='-'){
flag=-1; N=gc(); break;
}
ret = 0;
do{
ret = (ret<<3) + (ret<<1) + (N & 0b1111); N=gc();
}while('0'<=N&&N<='9');
ret *= flag;
}

int fastreadINT_byValue(){
int N = gc(), ret = 0, flag = 1;
for(;N<'0'||N>'9';N=gc())
if(N=='-'){
flag=-1; N=gc(); break;
}
do{
ret = (ret<<3) + (ret<<1) + (N & 0b1111); N=gc();
}while('0'<=N&&N<='9');
return ret*flag;
}

long long solutions(int &m,vector<long long> &times){
sort(times.begin(),times.end());
long long left = 1, answer, mid, sum;
long long right = answer = times.back()*m;
while(left+1<right){
mid = (left+right)>>1;
sum = 0;
for(auto it : times)
sum += (mid / it);

if(sum>=m) {answer = min(answer,mid); right = mid;}
else left = mid;
}
return answer;
}
int main(int argc, char** argv){
int tc, T, N, M;
vector<long long> times;
times.reserve(100001);
FIR(T);
for(tc = 1; tc <= T; tc++){
FIR(N); FIR(M);
times.clear();
for(int i = 0; i < N; i++)
times.push_back((long long)FIV());
printf("#%d %ld\n",tc,solutions(M,times));
}
return 0;//정상종료시 반드시 0을 리턴해야합니다.
}
Author: Song Hayoung
Link: https://songhayoung.github.io/2020/08/04/PS/SWEA/3074/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.