Hdu 1059 Dividing -- 多重背包

/*http://acm.hdu.edu.cn/showproblem.php?pid=1059  Dividing多重背包, 六种质量(1,2,3,4,5,6)的大理石，每种n[i]长，是否能分割为等质量的两份大理石，因为格式问题纠结死了*/#include <cstdio>#include <iostream>#include <string>#include <cstring>#define CLR(c,v) (memset(c,v,sizeof(c)))using namespace std;const int INF =   1<<30 ;const int inf = -(1<<30);const int M   = 21e4 + 10; // 21万int n[10]; // 物品的个数int dp[M];void CompletePack(int cost , int value, int max_cost){for (int i = cost ; i <= max_cost ; i++){if(dp[i] < dp[i-cost] + value){dp[i] = dp[i-cost] + value;}}}void ZeroOnePack(int cost ,int value, int max_cost){for (int i = max_cost ; i >= cost ; i--){if (dp[i] < dp[i-cost] + value){dp[i] = dp[i-cost] + value;}}}void MultiplePack(int cost ,int value,int amount, int max_cost){if (max_cost < cost*amount){ // 如果总数过多CompletePack(cost , value , max_cost);}else{for (int k = 1 ; k <= amount ; k <<= 1 ){ // 二进制优化, 注意k <<= 1 总是忘记等号if (k*cost <= max_cost){ZeroOnePack(cost*k , value*k ,max_cost);amount -= k;}}ZeroOnePack(cost*amount , value*amount , max_cost); // 剩余}}int main(){//freopen("in.txt","r",stdin);int Ncase = 0;while(cin >> n[0] >> n[1] >> n[2] >> n[3] >> n[4] >> n[5] && (n[0]||n[1]||n[2]||n[3]||n[4]||n[5]) ){CLR(dp,0);int sum = 0;for (int i = 1 ; i <= 6 ; i++){sum += n[i-1] * i; }if (sum & 1){printf("Collection #%d:\nCan't be divided.\n\n",++Ncase);continue;}else{for (int i = 1 ; i <= 6 ; i++){MultiplePack(i , i , n[i-1] , sum >> 1);}}if (sum >> 1 == dp[sum >> 1]){printf("Collection #%d:\nCan be divided.\n\n",++Ncase);}else{printf("Collection #%d:\nCan't be divided.\n\n",++Ncase);}}return 0;}