hdu 1059 Dividing

DividingTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8874    Accepted Submission(s): 2407Problem DescriptionMarsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection inhalf. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets thesame total value. Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into setsof equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles. InputEach line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 20 0''. The maximum total number of marbles will be 20000. The last line of the input file will be ``0 0 0 0 0 0''; do not process this line. OutputFor each colletcion, output ``Collection #k:'', where k is the number of the test case, and then either ``Can be divided.'' or ``Can't be divided.''. Output a blank line after each test case. Sample Input

1 0 1 2 0 01 0 0 0 1 10 0 0 0 0 0 Sample Output

Collection #1:Can't be divided.Collection #2:Can be divided. SourceMid-Central European Regional Contest 1999 RecommendJGShining

#include <stdio.h>int dp[120005];int V,v;//01背包, c表示花费, w表示价值void bag01(int c,int w) {for(v=V;v>=c;v--)if(dp[v]<dp[v-c]+w)dp[v]=dp[v-c]+w;}//完全背包,c表示花费, w表示价值void bagall(int c,int w) {for(v=c;v<=V;v++)if(dp[v]<dp[v-c]+w)dp[v]=dp[v-c]+w;}//多重背包,c表示花费, w表示价值,p表示个数void mutibag(int c,int w,int p) {if(c*p>=V)bagall(c,w);else {int k=1;while(k<p) {bag01(c*k,w*k);p-=k;k+=k;}bag01(c*p,w*p);}}int main(){int n[8];int i;int sum;int p=0;while(scanf("%d%d%d%d%d%d",&n[1],&n[2],&n[3],&n[4],&n[5],&n[6]),n[1]+n[2]+n[3]+n[4]+n[5]+n[6]) {sum=n[1]+n[2]*2+n[3]*3+n[4]*4+n[5]*5+n[6]*6;//sum为奇数个，那么肯定不能平分if(sum%2) {printf("Collection #%d:/nCan't be divided./n/n",++p);continue;}V=sum/2;    // 对总数的一半进行背包for(i=0;i<=V;i++) dp[i]=0;   //初始化for(i=1;i<=6;i++) mutibag(i,i,n[i]);  //进行多次背包if(dp[V]==V)printf("Collection #%d:/nCan be divided./n/n",++p);elseprintf("Collection #%d:/nCan't be divided./n/n",++p);}}