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这里;http://acm.hdu.edu.cn/showproblem.php?pid=1059
题意是有价值分别为1,2,3,4,5,6的商品各若干个,给出每种商品的数量,问是否能够分成价值相等的两份.
联想到多重背包,稍微用二进制优化一下。(最近身体不适,压力山大啊)
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #define inf 70000 5 using namespace std; 6 int dp[inf]; 7 int sum; 8 void pack(int price) 9 { 10 for(int i = sum; i >= price; i--) dp[i] = max(dp[i], dp[i - price] + price); 11 } 12 int main() 13 { 14 int a[8],i,q=1,j; 15 while (~scanf("%d %d %d %d %d %d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])) 16 { 17 if (a[1]==0&&a[2]==0&&a[3]==0&&a[4]==0&&a[5]==0&&a[6]==0) 18 break; 19 memset(dp,-inf,sizeof(dp)); 20 dp[0]=0; 21 printf("Collection #%d:\n",q++); 22 sum=0; 23 for (i=1;i<=6;i++) 24 sum+=a[i]*i; 25 if (sum%2!=0) 26 { 27 printf("Can‘t be divided.\n\n"); 28 continue; 29 } 30 sum/=2; 31 for (i=1;i<=6;i++) 32 { 33 if (i*a[i]>=sum) 34 { 35 for (j=i;j<=sum;j++) 36 dp[j]=max(dp[j],dp[j-i]+i); 37 } 38 else 39 { 40 int k = 1; 41 while(k < a[i]) 42 { 43 pack(k * i); 44 a[i] -= k; 45 k += k; 46 } 47 pack(a[i] * i); 48 } 49 } 50 if (dp[sum]==sum) 51 printf("Can be divided.\n\n"); 52 else 53 printf("Can‘t be divided.\n\n"); 54 } 55 return 0; 56 }
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原文地址:http://www.cnblogs.com/JJCHEHEDA/p/4893049.html