Codeforces 788C The Great Mixing

The Great Mixing

化简一下公式后发现， 问题变成了， 取最少多少数能使其和为1, bitset优化一下背包就好啦。

题解中介绍了一种bfs的方法没， 感觉比较巧妙。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 1000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, k, a[N];
bitset<2010> dp[N];

int main() {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= k; i++) scanf("%d", &a[i]);
    for(int i = 1; i <= k; i++) {
        if(a[i] == n) {
            puts("0");
            return 0;
        }
        a[i] -= n;
    }
    sort(a + 1, a + 1 + k);
    k = unique(a + 1, a + 1 + k) - a - 1;
    dp[0][1000] = 1;
    for(int i = 1; i <= 1000; i++) {
        for(int j = 1; j <= k; j++) {
            if(a[j] >= 0) dp[i] |= dp[i - 1] << a[j];
            else dp[i] |= dp[i - 1] >> (-a[j]);
            if(dp[i][1000]) {
                printf("%d\n", i);
                return 0;
            }
        }
    }
    puts("-1");
    return 0;
}

/*
*/