POJ

A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.

Input

The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.

Output

Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.

Sample Input

310090200

Sample Output

190 200

题意：给你一些数字。分成两组，这两组差得最小。样例：100+90=190 与200相差10 差最小。

#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <cstdio>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <time.h>using namespace std;#define pi acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x) memset(x,0,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0);typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;//const int dx[]={-1,0,1,0,-1,-1,1,1};//const int dy[]={0,1,0,-1,1,-1,1,-1};const int maxn=1e3+5;const int maxx=1e6+100;const double EPS=1e-7;const int MOD=1000000007;#define mod(x) ((x)%MOD);template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}//typedef tree<pt,null_type,less< pt >,rb_tree_tag,tree_order_statistics_node_update> rbtree;/*lch[root] = build(L1,p-1,L2+1,L2+cnt);    rch[root] = build(p+1,R1,L2+cnt+1,R2);中前*//*lch[root] = build(L1,p-1,L2,L2+cnt-1);    rch[root] = build(p+1,R1,L2+cnt,R2-1);中后*/long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);}int a[110];int main(){    int n;    int i,j,k;    srand();    while(scanf("%d",&n)!=EOF)    {        int mid=n/2;        int ans1=0,ans2=0;        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            if(i<=mid)ans1+=a[i];            else ans2+=a[i];        }        if(n==1){printf("%d %d\n",a[1],a[1]);continue;}        int minm=abs(ans1-ans2);        for(i=1;i<=50000;i++)        {            int l=(rand()%mid) + 1;            int r=(rand()%(n-mid)) + mid+1;            int t1=ans1-a[l]+a[r];            int t2=ans2-a[r]+a[l];            if(abs(t1-t2)<minm)            {                minm=abs(t1-t2);                ans1=t1;                ans2=t2;                int tmp=a[l];                a[l]=a[r];                a[r]=tmp;            }        }        if(ans1>ans2)swap(ans1,ans2);        printf("%d %d\n",ans1,ans2);    }    return 0;}