链接：点我
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 using namespace std; 6 const int v = 2000 + 5; 7 const int MaxN = 100 + 5; 8 int N, sum, num[MaxN], dp[2][v]; 9 int main() { 10 int i, j, k; 11 cin >> N; 12 for(i = 0; i < N; ++i) { 13 cin >> num[i]; 14 sum += num[i]; 15 } 16 memset(dp, -1, sizeof(dp)); 17 dp[1][0] = 0; 18 int a; 19 for(i = 0; i < N; ++i) { 20 a = i % 2; 21 memset(dp[a], -1, sizeof(dp[a])); 22 for(j = 0; j <= sum; ++j) { 23 //1.不放第i块水晶; 24 if(dp[a ^ 1][j] > -1) 25 dp[a][j] = dp[a ^ 1][j]; 26 //2.放进去后，高塔变矮塔（第i块放在矮塔上了）; 27 if(num[i] > j && dp[a ^ 1][num[i] - j] > -1) 28 dp[a][j] = max(dp[a][j], dp[a ^ 1][num[i] - j] + j); 29 //3.放进去后，高塔仍高（第i块放在矮塔上）; 30 if(j + num[i] <= sum && dp[a ^ 1][j + num[i]] > -1) 31 dp[a][j] = max(dp[a][j], dp[a ^ 1][j + num[i]]); 32 //4.放进去后，高塔更高（第i块放在高塔上）. 33 if(j >= num[i] && dp[a ^ 1][j - num[i]] > -1) 34 dp[a][j] = max(dp[a][j], dp[a ^ 1][j - num[i]] + num[i]); 35 } 36 } 37 if(dp[a][0] > 0) 38 cout << dp[a][0] << endl; 39 else 40 cout << "Impossible" << endl; 41 }