XTU 1250

http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1250

 

Super Fast Fourier Transform

Bobo has two sequences of integers { a1,a2,…,an} and { b1,b2,…,bm}. He would like to find

∑i=1n∑j=1m⌊|ai−bj|−−−−−−−√⌋.

 

Note that ⌊x⌋ denotes the maximum integer does not exceed x, and |x| denotes the absolute value of x.

Input

The input contains at most 30 sets. For each set:

The first line contains 2 integers n,m (1≤n,m≤105).

The second line contains n integers a1,a2,…,an.

The thrid line contains m integers b1,b2,…,bm.

(ai,bi≥0,a1+a2+⋯+an,b1+b2+…,bm≤106)

Output

For each set, an integer denotes the sum.

Sample Input

1 2
1
2 3
2 3
1 2
3 4 5

Sample Output

2
7

不得不说自己笨，这么简单，当时就是不会

题意：n个a,m个b的相乘的累积和

思路：把重复的a和b找出了，会减少很大的复杂度

 

 1 #include 
 2 #include 
 3 #include <string.h>
 4 #include 
 5 
 6 int a[100005],b[100005],suma[1000005],sumb[1000005];
 7 long long ans;
 8 
 9 int main()
10 {
11     int m,n,acut,bcut,tmp;
12     while(~scanf("%d%d",&m,&n))
13     {
14         acut = 0,bcut = 0;
15         memset(suma,0,sizeof(suma));
16         memset(sumb,0,sizeof(sumb));
17         for(int i = 1;i<=m;i++)
18         {
19             scanf("%d",&tmp);
20             if(suma[tmp]==0)
21                 a[acut++] = tmp;
22             suma[tmp]++;
23         }
24         for(int i = 1;i<=n;i++)
25         {
26             scanf("%d",&tmp);
27             if(sumb[tmp]==0)
28                 b[bcut++] = tmp;
29             sumb[tmp]++;
30         }
31         ans = 0;
32         for(int i = 0;i)
33             for(int j = 0;j)
34             {
35                 ans+=(long long ) suma[a[i]]*sumb[b[j]]*(long long )(sqrt(abs(a[i]-b[j])));
36             }
37         printf("%lld\n",ans);
38     }
39     return 0;
40 }

 

转载于:https://www.cnblogs.com/Tree-dream/p/6540096.html