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usaco Humble Numbers

2018年04月23日 ⁄ 综合 ⁄ 共 1342字 ⁄ 字号 评论关闭

这个模拟也是真心不会做啊,第一次做超时,然后看了标程..

code:

/*
    ID:yueqi
    LANG:C++
    TASK:humble
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits.h>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("humble.in","r",stdin)
#define writef freopen("humble.out","w",stdout)
const int maxn = 100001;
const long long BigP=999983;
const long long  INF = 0x5fffffff;
const int dx[]={-1,0,1,0};
const int dy[]={0,1,0,-1};
const double pi = acos(-1.0);
const double eps= 1e-7;
using namespace std;
LL dp[maxn];
int prime[101],pos[101];
int N,M,num;
int main(){
    readf;
    writef;
    SD(N);SD(M);
    FOR(i,1,N) SD(prime[i]);
    dp[0]=1;
    LL minn;
    int minnpos;
    while(num<M){
        minn=100*INF;
        minnpos=-1;
        FOR(i,1,N){
            while(prime[i]*dp[pos[i]]<=dp[num]) pos[i]++;
            if(prime[i]*dp[pos[i]]<minn){
                minn=prime[i]*dp[pos[i]];
                minnpos=i;
            }
        }
        dp[++num]=minn;
        pos[minnpos]++;//?
    }
    printf("%lld\n",dp[M]);
    return 0;
}

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