标签：des style blog class c code
Mary and Rose own a collection of jewells. They want to split the collection among themselves so that both receive an equal share of the jewels. This would be easy if all the jewels had the same value, because then they could just split the collection in half. But unfortunately, some of the jewels are larger, or more beautiful than others. So, Mary and Rose start by assigning a value, a natural number between one and ten, to each jewel. Now they want to divide the jewels so that each of them gets the same total value. Unfortunately, they realize that it might be impossible to divide the jewels in this way (even if the total value of all jewels is even). For example, if there are one jewel of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the jewels.
1 0 1 2 0 0 0 0 2 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#1:Can‘t be divided. #2:Can be divided.
1 #include<iostream> 2 #include<cstdio> 3 #include<string> 4 #include<string.h> 5 #include<algorithm> 6 #include<cmath> 7 #include<vector> 8 #include<cstring> 9 #include<stack> 10 #include<stdlib.h> 11 #include<ctype.h> 12 using namespace std; 13 #define MAXN 1300 14 #define inf 1000000 15 16 17 int a[11],dp[50001]; 18 int main() 19 { 20 int flag=1; 21 while(1) 22 { 23 int sum=0,k=0; 24 for(int i=1;i<11;i++) 25 { 26 cin>>a[i]; 27 sum+=(a[i]*i); 28 k+=a[i]; 29 } 30 if(k==0)break; 31 if(sum%2!=0){printf("#%d:Can‘t be divided.\n\n",flag++);continue;} 32 sum/=2; 33 for(int i=0;i<sum+1;i++) 34 dp[i]=0; 35 for(int i=1;i<11;i++) 36 for(int j=1;j<=a[i];j++) 37 for(int k=i;k<=sum;k++) 38 { 39 dp[k]=max(dp[k],dp[k-i]+j*i); 40 } 41 if(dp[sum]==sum) 42 printf("#%d:Can be divided.\n\n",flag++); 43 else 44 printf("#%d:Can‘t be divided.\n\n",flag++); 45 46 } 47 }
后来看看，发现这题怎么那么眼熟，看自己以前的博客（csdn）发现，确实是类型题，所以就改下数据，水过了！
表示这个可以当模板用啊！
代码：
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<map> 5 #include<vector> 6 #include<set> 7 #include<string> 8 #include<algorithm> 9 #include<cmath> 10 using namespace std; 11 #define Max(a,b) a>b?a:b 12 #define Min(a,b) a<b?a:b 13 #define inf 99999999 14 int dp[50020],num[11],sum; 15 16 void zero_one(int cost,int weight)//01背包 17 { 18 for(int j=sum;j>=cost;j--) 19 { 20 dp[j]=Max(dp[j],dp[j-cost]+weight); 21 22 } 23 } 24 void totle_pack(int cost,int weight)//完全背包 25 { 26 for(int j=0;j<=sum;j++) 27 { 28 if(j>=cost) 29 dp[j]=Max(dp[j],dp[j-cost]+weight); 30 } 31 } 32 void multiple_pack(int cost,int weight ,int n)//多重背包 33 { 34 if(n*cost>=sum) 35 totle_pack(cost,weight); 36 else 37 { 38 int k=1; 39 while(k<n) 40 { 41 zero_one(cost*k,weight*k); 42 n-=k; 43 k*=2; 44 } 45 zero_one(n*cost,n*weight); 46 } 47 48 } 49 50 int main() 51 { 52 int t=1; 53 while(1) 54 { 55 sum=0; 56 for(int i=1;i<=10;i++) 57 { 58 cin>>num[i]; 59 sum+=num[i]*i; 60 } 61 if(sum==0)break; 62 63 cout<<"#"<<(t++)<<‘:‘; 64 if(sum%2==1) 65 { 66 cout<<"Can‘t be divided."<<endl<<endl; 67 } 68 else 69 { 70 sum/=2; 71 for(int i=0;i<=sum;i++) 72 dp[i]=0; 73 for(int i=1;i<=10;i++) 74 { 75 if(num[i]) 76 multiple_pack(i,i,num[i]); 77 } 78 if(dp[sum]==sum) 79 cout<<"Can be divided."<<endl<<endl; 80 else 81 cout<<"Can‘t be divided."<<endl<<endl; 82 } 83 } 84 return 0; 85 }
Divideing Jewels（nyoj546）（多重背包+二进制优化）,布布扣,bubuko.com
Divideing Jewels（nyoj546）（多重背包+二进制优化）
标签：des style blog class c code
原文地址：http://www.cnblogs.com/zhaopengze/p/3739223.html
