hdu 1059 Dividing(多重背包)
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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 sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.
Output a blank line after each test case.
1 0 1 2 0 01 0 0 0 1 10 0 0 0 0 0
Collection #1:Can't be divided.Collection #2:Can be divided.
题意:给你1,2,3,4,5,6这6个价值的数量,问你能不能平分,当然每个价值是不能拆分的。
思路:多重背包啊 (不太了解多重背包的话,请看这道例题http://blog.csdn.net/qqchanmingdexiaji/article/details/53842460)
代码:
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n;int c[10]= {1,2,3,4,5,6},mi[10];int dp[120005];void Zp(int c,int w) { int i; for(i=n; i>=c; i--) dp[i]=max(dp[i],dp[i-c]+w);}void Cp(int c,int w) { int i; for(i=c; i<=n; i++) dp[i]=max(dp[i],dp[i-c]+w);}void Mp(int c,int w,int sum) { if(n<=c*sum) { Cp(c,w); return ; } else { int k=1; while(k<=sum) { Zp(k*c,k*w); sum-=k; k*=2; } Zp(sum*c,sum*w); }}int main() { int cas=1; int i; while(scanf("%d%d%d%d%d%d",&mi[0],&mi[1],&mi[2],&mi[3],&mi[4],&mi[5]) && mi[0]+mi[1]+mi[2]+mi[3]+mi[4]+mi[5]) { memset(dp,0,sizeof(dp)); printf("Collection #%d:\n",cas); cas++; n=0; for(i=0; i<6; i++) n+=c[i]*mi[i]; if(n%2!=0) printf("Can't be divided.\n\n"); else { n/=2; for(i=0; i<6; i++) Mp(c[i],c[i],mi[i]); if(dp[n]==n) printf("Can be divided.\n\n");//输出也要注意 else printf("Can't be divided.\n\n"); } }}
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