51 nod 1007 正整数分组 (简单01背包)

http://www.51nod.com/onlineJudge/questionCode.html#problemId=1007&noticeId=15020

求出n个数的和sum,然后用sum/2作为背包容量,让n个数去放,求出一个最大价值,那么这就是其中一组的和,另外一组的和就是sum-dp[sum/2];

注意这里的体积和价值都是a[i];

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <vector>
 5 #include <cstring>
 6 #include <string>
 7 #include <algorithm>
 8 #include <string>
 9 #include <set>
10 #include <functional>
11 #include <numeric>
12 #include <sstream>
13 #include <stack>
14 #include <map>
15 #include <queue>
16 #pragma comment(linker, "/STACK:102400000,102400000")
17 #define CL(arr, val)    memset(arr, val, sizeof(arr))
18
19 #define ll long long
20 #define inf 0x7f7f7f7f
21 #define lc l,m,rt<<1
22 #define rc m + 1,r,rt<<1|1
23 #define pi acos(-1.0)
24
25 #define L(x)    (x) << 1
26 #define R(x)    (x) << 1 | 1
27 #define MID(l, r)   (l + r) >> 1
28 #define Min(x, y)   (x) < (y) ? (x) : (y)
29 #define Max(x, y)   (x) < (y) ? (y) : (x)
30 #define E(x)        (1 << (x))
31 #define iabs(x)     (x) < 0 ? -(x) : (x)
32 #define OUT(x)  printf("%I64d\n", x)
33 #define lowbit(x)   (x)&(-x)
34 #define Read()  freopen("a.txt", "r", stdin)
35 #define Write() freopen("b.txt", "w", stdout);
36 #define maxn 1000000000
37 #define N 2510
38 #define mod 1000000000
39 using namespace std;
40
41 int a[10001],dp[10001];
42 int main()
43 {
44    // freopen("a.txt","r",stdin);
45     int n,sum=0,n1=0;
46     scanf("%d",&n);
47     for(int i=0;i<n;i++)
48     {
49         scanf("%d",&a[i]);
50         sum+=a[i];
51     }
52     n1=sum/2;
53     memset(dp,0,sizeof(dp));
54     for(int i=n-1;i>=0;i--)
55        for(int j=n1;j>=0;j--)
56        {
57            if(j>=a[i])
58            {
59                dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
60            }
61        }
62     //printf("%d\n",dp[n1]);
63     printf("%d\n",abs(sum-dp[n1]-dp[n1]));
64     return 0;
65 }