HDOJ 1709 The Balance(母函数)
2016-01-29 01:20
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Total Submission(s): 7060 Accepted Submission(s): 2912
[align=left]Problem Description[/align]
Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range
[1,S]. S is the total quality of all the weights.
[align=left]Input[/align]
The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N),
indicating the quality of each weight where 1<=Ai<=100.
[align=left]Output[/align]
For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not
zero.
[align=left]Sample Input[/align]
3
1 2 4
3
9 2 1
[align=left]Sample Output[/align]
0
2
4 5
题意: 有一个天平和n个砝码,给出这些砝码的质量ai,现在问你在[1,s](s表示砝码质量之和)间有哪些质量是秤不出来的,先输出个数,在输出确定的数。
题解:尴尬,这么简单的题都没读懂。 母函数变形啦,砝码是放在天平的左右两边的,对于每种质量我们可以ai+aj==weight,也可以ai-aj==weight。 所以在进行合并乘法式是注意 c2[k+j]+=c1[j], 也有c2[abs(j-k)]+=c1[j]。
代码如下:
The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7060 Accepted Submission(s): 2912
[align=left]Problem Description[/align]
Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range
[1,S]. S is the total quality of all the weights.
[align=left]Input[/align]
The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N),
indicating the quality of each weight where 1<=Ai<=100.
[align=left]Output[/align]
For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not
zero.
[align=left]Sample Input[/align]
3
1 2 4
3
9 2 1
[align=left]Sample Output[/align]
0
2
4 5
题意: 有一个天平和n个砝码,给出这些砝码的质量ai,现在问你在[1,s](s表示砝码质量之和)间有哪些质量是秤不出来的,先输出个数,在输出确定的数。
题解:尴尬,这么简单的题都没读懂。 母函数变形啦,砝码是放在天平的左右两边的,对于每种质量我们可以ai+aj==weight,也可以ai-aj==weight。 所以在进行合并乘法式是注意 c2[k+j]+=c1[j], 也有c2[abs(j-k)]+=c1[j]。
代码如下:
#include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #define maxn 10010 int c1[maxn],c2[maxn]; int a[110]; int main() { int n,i,j,k,cnt,sum,num; while(scanf("%d",&n)!=EOF) { sum=0; for(i=1;i<=n;++i) { scanf("%d",&a[i]); sum+=a[i]; } memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); c1[0]=c1[a[1]]=1; num=a[1]; for(i=2;i<=n;++i) { for(j=0;j<=num;++j) { for(k=0;k<=a[i]&&k+j<=sum;k+=a[i]) { c2[k+j]+=c1[j]; c2[abs(j-k)]+=c1[j]; } } num+=a[i]; for(j=0;j<=num;++j) { c1[j]=c2[j]; c2[j]=0; } } int cnt=0; for(i=0;i<=sum;++i) { if(!c1[i]) c2[cnt++]=i; } if(cnt==0) printf("0\n"); else { printf("%d\n",cnt); for(i=0;i<cnt;++i) { if(i==cnt-1) printf("%d\n",c2[i]); else printf("%d ",c2[i]); } } } return 0; }
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