[hdu3631]背包或中途相遇法

 1 #pragma comment(linker, "/STACK:10240000,10240000")
 2 
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <cstdlib>
 7 #include <cstring>
 8 #include <map>
 9 #include <queue>
10 #include <deque>
11 #include <cmath>
12 #include <vector>
13 #include <ctime>
14 #include <cctype>
15 #include <set>
16 
17 using namespace std;
18 
19 #define mem0(a) memset(a, 0, sizeof(a))
20 #define lson l, m, rt << 1
21 #define rson m + 1, r, rt << 1 | 1
22 #define define_m int m = (l + r) >> 1
23 #define rep(a, b) for(int a = 0; a < b; a++)
24 #define all(a) (a).begin(), (a).end()
25 #define lowbit(x) ((x) & (-(x)))
26 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
27 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
28 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
29 #define pc(a) putchar(a)
30 #define ps(a) printf("%s", a)
31 #define pd(a) printf("%d", a)
32 #define sd(a) scanf("%d", &a)
33 
34 typedef double db;
35 typedef long long LL;
36 typedef pair<int, int> pii;
37 typedef multiset<int> msi;
38 typedef set<int> si;
39 typedef vector<int> vi;
40 typedef map<int, int> mii;
41 
42 const int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
43 const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
44 const int maxn = 1e5 + 7;
45 const int maxm = 1e5 + 7;
46 const int maxv = 1e7 + 7;
47 const int max_val = 1e6 + 7;
48 const int MD = 1e9 +7;
49 const int INF = 1e9 + 7;
50 const double PI = acos(-1.0);
51 const double eps = 1e-10;
52 
53 template<class T> T gcd(T a, T b) { return b == 0? a : gcd(b, a % b); }
54 
55 int t, sum[33], a[33];
56 bool f[10000007];
57 int main() {
58     //freopen("in.txt", "r", stdin);
59     int n;
60     while (cin >> n >> t) {
61         rep(i, n) sd(a[i]);
62         sort(a, a + n);
63         sum[0] = a[0];
64         rep(i, n - 1) sum[i + 1] = sum[i] + a[i + 1];
65         mem0(f);
66         f[0] = true;
67         rep(i, n) {
68             int maxt = min(t, sum[i]);
69             for (int j = maxt; j >= a[i]; j--) {
70                 f[j] = f[j] || f[j - a[i]];
71             }
72         }
73         int ans;
74         for (int i = t; i >= 0; i--) {
75             if (f[i]) {
76                 ans = i;
77                 break;
78             }
79         }
80         cout << ans << endl;
81     }
82     return 0;
83 }