CodeForces

632E：http://codeforces.com/problemset/problem/632/E

参考：https://blog.csdn.net/qq_21057881/article/details/51023067

题意：

　　　　给定n个值，让你选择k个数，可以重复选择，问可以得到哪些数字。

思路：

　　　　显然最小的值起到很大的作用，我们可以把每个值都减去这个最小值，利用完全背包，建立dp【i】表示，取到 i 这么多值最少需要多少个数。如果取到 i值需要的数值小于等于K，那么k * 最小值 + i 就是我们能取到的一个值，因为一定需要 k*最小值 （注意我们一开始让每个数都剪去了最小值，要加回来）。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
// const int mod = 10007;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/

            const int maxn = 1009;
            int a[maxn],dp[maxn*maxn];
int main(){
            int n,k;
            scanf("%d%d", &n, &k);
            for(int i=1; i<=n; i++)scanf("%d", &a[i]);
            sort(a+1,a+1+n);

            n = unique(a+1,a+1+n) - (a+1);
            // debug(n);
            for(int i=2; i<=n; i++) a[i] = a[i] - a[1];

            memset(dp,inf,sizeof(dp));
            dp[0] = 0;
            for(int i=2; i<=n; i++){
                for(int j=a[i]; j<=a[i]*k; j++){
                    dp[j] = min(dp[j] , dp[j-a[i]] + 1);
                }
            }
            for(int i=0; i<=a[n] * k; i++){
                if(dp[i] <= k){
                    printf("%d ", a[1] * k + i);
                }
            }
            printf("
");
            return 0;
}