1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=503
2、题目大意：
给定一个背包，里边有m件硬币，每种硬币的价值给出，现在要将这些硬币分给两个人，让两个人的金钱总数差值尽可能的小，用01背包就能解决，背包较小的那个的最大价值为背包容量的一半
3、题目：
Dividing coins
| Dividing coins |
It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great length and thus created copper-wire.
Not commonly known is that the fighting started, after the two Dutch tried to divide a bag with coins between the two of them. The contents of the bag appeared not to be equally divisible. The Dutch of the past couldn't stand the fact that a division should favour one of them and they always wanted a fair share to the very last cent. Nowadays fighting over a single cent will not be seen anymore, but being capable of making an equal division as fair as possible is something that will remain important forever...
That's what this whole problem is about. Not everyone is capable of seeing instantly what's the most fair division of a bag of coins between two persons. Your help is asked to solve this problem.
Given a bag with a maximum of 100 coins, determine the most fair division between two persons. This means that the difference between the amount each person obtains should be minimised. The value of a coin varies from 1 cent to 500 cents. It's not allowed to split a single coin.
Input
A line with the number of problems n, followed by n times:
- a line with a non negative integer m () indicating the number of coins in the bag
- a line with m numbers separated by one space, each number indicates the value of a coin.
Output
The output consists of n lines. Each line contains the minimal positive difference between the amount the two persons obtain when they divide the coins from the corresponding bag.
Sample Input
2 3 2 3 5 4 1 2 4 6
Sample Output
0 1
4、AC的二维数组的代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 105
int v[N];
int dp[N][25105];
int main()
{int t,m,sum;while(scanf("%d",&t)!=EOF){while(t--){sum=0;scanf("%d",&m);for(int i=1; i<=m; i++){scanf("%d",&v[i]);sum+=v[i];}int ans=sum/2;memset(dp,0,sizeof(dp));for(int i=1; i<=m; i++){for(int j=0; j<=ans; j++){
// if(j-v[i]>=0)
// dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+v[i]);
//这样写是错的，如果当期物品不能装入，那么dp[i][j]就不能选择不装的情况，
//因为if进不去，dp[i][j]就不能得到更新dp[i][j]=dp[i-1][j];if(j-v[i]>=0)dp[i][j]=max(dp[i][j],dp[i-1][j-v[i]]+v[i]);}}//printf("*%d\n",dp[m][ans]);int res=sum-2*dp[m][ans];printf("%d\n",res);}}return 0;
}
AC的一维数组的代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 105
int v[N];
int dp[25005];
int main()
{int t,m,sum;scanf("%d",&t);while(t--){sum=0;scanf("%d",&m);for(int i=1;i<=m;i++){scanf("%d",&v[i]);sum+=v[i];}int ans=sum/2;memset(dp,0,sizeof(dp));for(int i=1;i<=m;i++){for(int j=ans;j>=0;j--){if(j>=v[i])dp[j]=max(dp[j],dp[j-v[i]]+v[i]);}}//printf("*%d\n",dp[m][ans]);int res=abs(sum-dp[ans]-dp[ans]);printf("%d\n",res);}return 0;
}
/*
2
3
2 3 5
4
1 2 4 6
*/
4、代码：二维的dp [][] wrong answer了
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 105
int v[N];
int dp[N][50005];
int main()
{int t,m,sum;scanf("%d",&t);while(t--){sum=0;scanf("%d",&m);for(int i=1;i<=m;i++){scanf("%d",&v[i]);sum+=v[i];}int ans=sum/2+1;memset(dp,0,sizeof(dp));for(int i=1;i<=m;i++){for(int j=0;j<=ans;j++){if(j-v[i]>=0)dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+v[i]);}}//printf("*%d\n",dp[m][ans]);int res=abs(sum-dp[m][ans]-dp[m][ans]);printf("%d\n",res);}return 0;
}
改成一维的dp也wrong answer了
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 105
int v[N];
int dp[50005];
int main()
{int t,m,sum;scanf("%d",&t);while(t--){sum=0;scanf("%d",&m);for(int i=1;i<=m;i++){scanf("%d",&v[i]);sum+=v[i];}int ans=sum/2+1;memset(dp,0,sizeof(dp));for(int i=1;i<=m;i++){for(int j=ans;j>=0;j--){if(j-v[i]>=0)dp[j]=max(dp[j],dp[j-v[i]]+v[i]);}}//printf("*%d\n",dp[m][ans]);int res=abs(sum-dp[ans]-dp[ans]);printf("%d\n",res);}return 0;
}
/*
2
3
2 3 5
4
1 2 4 6
*/