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void multbag(int c,int w,int n)//多重背包 { if(c*n>=V)//当某一项的价值总和比需要求的价值综合还更大时可以看作完全背包问题 { bagall(c,w);return ; } int k=1; while(k<=n)//不然就看做很多个0-1背包问题的总和 { bag01(k*c,k*w); n=n-k; k=k*2; } bag01(n*c,n*w...
Dividing Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25778 Accepted Submission(s): 7379 Problem ...HDU 1059 Dividing(多重背包 , 二进制优化形成01背包) 题目链接: 1059 Dividing 题意: 有6个物品,一个物品的价值=该物品的编号i (1-...
Dividing Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14505 Accepted Submission(s): 4064 Problem Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal...
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 1601 Accepted Submission(s): 687 题目链接:点击打开链接 Problem Description RXD has a treeT, with the size ofn. Each edge has a cost. ...
import java.util.Scanner; public class Main{ private static Scanner scanner; public static void main(String[] args) { scanner = new Scanner(System.in); int count = 1; while (scanner.hasNext()) { int arr[] = new int[6]; for (int i = 0; i < arr.length; i++) { ...
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int main() { int arr[] = { 2, 5 , 8, 11, 43 , 78 , 234 }; int n = sizeof(arr)/sizeof(arr[0]); int sum = arr[0]; for (int i = 1; i < n; i++) { sum += arr[i] / arr[i - 1]; } printf("The sum of array after dividing number from previous numbers is...
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 113 Accepted Submission(s): 50 Problem Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of ...
(System.in);String s=scanner.nextLine();int[]N=newint[6];int collection=1;while(!s.equals(ENDING_LINE)){int pack=0;String[]t=s.split(" ");for(int i=0;i<6;++i){N[i]=Integer.parseInt(t[i]);pack+=(i+1)*N[i];}if((1&pack)==1)outPutResult(DIVIDE_FALSE,collection);...