szoj461【四校联考0430】挑战

传送门：（涉及版权忽略）

【题解】

我们发现n的范围很小，提示我们可以折半，然后我们就会了O(T2^(n/2)*n)的做法，然而会T。

考虑如何优化。直接排序会多一个log(2^(n/2))也就是n，那么改成每次加一个数，归并即可。这样复杂度是对的

T(n) = T(n-1) + 2^n  ==> T(n) = O(2^n)

那么复杂度就是O(T2^(n/2))

# include <stdio.h>
# include <string.h>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M =  + , N = ;
const int mod = 1e9+;

# define FO_OPEN
# define RG register
# define ST static

int n, m, a[M];
int c[][N], cn[];
int t[N];

inline void merge(int pos, int l, int mid, int r) {
    int i = l, j = mid+, k = l-;
    while(i<=mid && j<=r) {
        if(c[pos][i] < c[pos][j]) t[++k] = c[pos][i++];
        else t[++k] = c[pos][j++];
    }
    while(i<=mid) t[++k] = c[pos][i++];
    while(j<=r) t[++k] = c[pos][j++];
    for (int o=l; o<=r; ++o) c[pos][o] = t[o];
}

inline void sol() {
    int sum = ;
    scanf("%d%d", &n, &m);
    for (int i=; i<=n; ++i) scanf("%d", &a[i]), sum = sum + a[i];
    if(sum < m) {
        puts("-1");
        return ;
    }
    int res = n/;
    c[][cn[] = ] = ;
    for (int i=; i<=res; ++i) {
        for (int j=; j<=cn[]; ++j) c[][j+cn[]] = c[][j] + a[i];
        merge(, , cn[], cn[] << );
        cn[] <<= ;
    }
//    printf("cn[0] = %d\n", cn[0]);
//    for (int i=1; i<=cn[0]; ++i) printf("%d ", c[0][i]);
//    puts("\n====================");
    c[][cn[] = ] = ;
    for (int i=res+; i<=n; ++i) {
        for (int j=; j<=cn[]; ++j) c[][j+cn[]] = c[][j] + a[i];
        merge(, , cn[], cn[] << );
        cn[] <<= ;
    }
//    printf("cn[1] = %d\n", cn[1]);
//    for (int i=1; i<=cn[1]; ++i) printf("%d ", c[1][i]);
//    puts("");
    int p0 = , p1 = cn[], ans = ;
    for (; p0 <= cn[]; p0 ++) {
        while(p1 && c[][p0] + c[][p1] >= m) --p1;
        if(p1 != cn[])
            ans = min(ans, c[][p0] + c[][p1 + ]);
    }
    printf("%d\n", ans);
} 

int main() {
    FO_OPEN ? freopen("challenge.in", "r", stdin), freopen("challenge.out", "w", stdout) : ;
    int T; scanf("%d", &T);
    while(T--) sol();
    return ;
}