UVA 562 Dividing coins(01背包)
2016-03-30 14:58
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题目链接:
UVA 562 Dividing coins
题意:
给出n个数,将这n个数分成两堆,使这两堆和尽可能接近,输出两堆和之差的绝对值。
分析:
记n个数的和为sum,则其中一对和一定是<=sum/2,而另一堆和>=sum/2。
那么就可以转化成这个问题:由这n个数组成不大于sum/2的最大数是多少?
UVA 562 Dividing coins
题意:
给出n个数,将这n个数分成两堆,使这两堆和尽可能接近,输出两堆和之差的绝对值。
分析:
记n个数的和为sum,则其中一对和一定是<=sum/2,而另一堆和>=sum/2。
那么就可以转化成这个问题:由这n个数组成不大于sum/2的最大数是多少?
//0K 13MS #include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <cmath> using namespace std; const int MAX_N=110; const int MAX_SUM=50010; int T,n,sum; int val[MAX_N],dp[MAX_SUM]; int main() { //freopen("uva562in.txt","r",stdin); scanf("%d",&T); while(T--){ scanf("%d",&n); sum=0; for(int i=0;i<n;i++){ scanf("%d",&val[i]); sum+=val[i]; } memset(dp,0,sizeof(dp));//dp[i]表示在总和不超过i的情况下能获得的最大总和 for(int i=0;i<n;i++){ for(int j=sum/2;j>=val[i];j--){ dp[j]=max(dp[j],dp[j-val[i]]+val[i]); } } //printf("sum=%d dp[sum/2]=%d\n",sum,dp[sum/2]); printf("%d\n",sum-2*dp[sum/2]);//(sum-dp[sum/2])-(dp[sum/2]) } return 0; }
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