Tug of War
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 8085 | Accepted: 2174 |
Description
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
3 100 90 200
Sample Output
190 200
Source
/* * Author: * Created Time: 2013/10/14 22:43:39 * File Name: A.cpp * solve: A.cpp */ #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> #include<string> #include<map> #include<stack> #include<set> #include<iostream> #include<vector> #include<queue> //ios_base::sync_with_stdio(false); //#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; #define sz(v) ((int)(v).size()) #define rep(i, a, b) for (int i = (a); i < (b); ++i) #define repf(i, a, b) for (int i = (a); i <= (b); ++i) #define repd(i, a, b) for (int i = (a); i >= (b); --i) #define clr(x) memset(x,0,sizeof(x)) #define clrs( x , y ) memset(x,y,sizeof(x)) #define out(x) printf(#x" %d ", x) #define sqr(x) ((x) * (x)) typedef long long LL; const int INF = 1000000000; const double eps = 1e-8; const int maxn = 300; int sgn(const double &x) { return (x > eps) - (x < -eps); } int num[maxn]; int dp[100+1][450*100+10];//dp[i][j]代表能否用i个数组成j int main() { //freopen("in.txt","r",stdin); int n; while(scanf("%d",&n)==1) { int sum = 0; repf(i,1,n) { scanf("%d",&num[i]); sum += num[i]; } clr(dp); dp[0][0] = 1; int N = (n+1)/2; repf(i,1,n) { repd(j,sum/2,0) repd(k,N,1) { if(j>=num[i]) { if(dp[k - 1][j - num[i]] == 1) dp[k][j] = 1; } } } int ans; repd(i,sum/2,0) { if(n%2 == 0) { if(dp[n/2][i] == 1) { ans = i; break; } }else { if(dp[n/2][i] == 1 || dp[n/2 + 1][i] == 1) { ans = i; break; } } } cout<<min(ans,sum-ans)<<" "<<max(ans,sum - ans)<<endl; } return 0; }
/* * Author: * Created Time: 2013/10/14 22:43:39 * File Name: A.cpp * solve: A.cpp */ #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> #include<string> #include<map> #include<stack> #include<set> #include<iostream> #include<vector> #include<queue> //ios_base::sync_with_stdio(false); //#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; #define sz(v) ((int)(v).size()) #define rep(i, a, b) for (int i = (a); i < (b); ++i) #define repf(i, a, b) for (int i = (a); i <= (b); ++i) #define repd(i, a, b) for (int i = (a); i >= (b); --i) #define clr(x) memset(x,0,sizeof(x)) #define clrs( x , y ) memset(x,y,sizeof(x)) #define out(x) printf(#x" %d ", x) #define sqr(x) ((x) * (x)) typedef long long LL; const int INF = 1000000000; const double eps = 1e-8; const int maxn = 300; int sgn(const double &x) { return (x > eps) - (x < -eps); } int num[maxn]; int dp[100+1][450*100+10];//dp[i][j]代表能否用i个数组成j int main() { //freopen("in.txt","r",stdin); int n; while(scanf("%d",&n)==1) { int sum = 0; repf(i,1,n) { scanf("%d",&num[i]); sum += num[i]; } clr(dp); dp[0][0] = 1; int N = (n+1)/2; repf(i,1,n) { repf(j,0,sum/2) repf(k,0,N) { if(dp[k][j] == 1) { if(j+num[i] <= sum/2) { dp[k+1][j+num[i]] = 1; } } } } int ans; repd(i,sum/2,0) { if(n%2 == 0) { if(dp[n/2][i] == 1) { ans = i; break; } }else { if(dp[n/2][i] == 1 || dp[n/2 + 1][i] == 1) { ans = i; break; } } } cout<<min(ans,sum-ans)<<" "<<max(ans,sum - ans)<<endl; } return 0; }