面试题：数组划分成两个子数组，和的差值的绝对值最小

#include <iostream>
#include <string>
#include <memory.h>
#include <vector>
using namespace std;

//给定一个数组，分成两组后，求两组的和的差的绝对值最小。
//元素值大于0，没有负数；元素个数小于100
/*
思路：背包问题，求最接近sum/2的子数组即可。
*/
int dp[101][10001];
int fun(vector<int> &arr)
{
int sum = 0;
int n = arr.size();
int i;
for(i = 0 ; i < n ; ++i)
{
sum += arr[i];
}
int sumorg = sum;
sum = sum/2;
for(i = 0 ; i <= sum ; ++i)
{
dp[0][i] = 0;
}
int j;
for(i = 1 ; i <= n ; ++i)
{
for(j = arr[i-1]; j <= sum ; ++j)
{
dp[i][j] = max(dp[i][j],dp[i-1][j-arr[i-1]+arr[i-1]);
}
}
return abs(sumorg,dp
[sum]);
}

int main()
{
return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>                  //for memset
#include <algorithm>                //for max
using namespace std;
const int M = 100 + 10;
int w[M];
int dp[M*100];
bool state[M][M];
int main()
{
int n;
while (scanf("%d", &n) != EOF) {
int sum = 0;
for (int i = 0; i < n; ++i) {
scanf("%d", &w[i]);
sum += w[i];
}
memset(dp, 0, sizeof(dp));
memset(state, 0, sizeof(state));
for (int i = 0; i < n; ++i)
for (int j = sum/2; j >= w[i]; --j) {
if (dp[j] < dp[j-w[i]] + w[i]) {
dp[j] = dp[j-w[i]] + w[i];
state[i][j] = true;
}
}
printf("%d\n", sum - dp[sum/2]*2);
int i = n, j = sum/2;
while (i--) {
if (state[i][j]) {
printf("%d ", w[i]);
j -= w[i];
}
}
printf("\n");
}
return 0;
}