2016长城信息杯中国大学生程序设计竞赛中南邀请赛 xtu 1250 Super Fast Fourier Transform

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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

贡献了十几波wa。。。。g++和g++11 编译器的问题？g++交了十几次 换个g++11就过了 omg

有10^5个数 然后所有数的和小于10^6次方 一眼感觉下去会有很多重复的数 去一下重就好了。。

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<string.h>
using namespace std;
int a[100005],b[100005],suma[1000005],sumb[1000005];
int main()
{
    int la,lb,t,n,m;
    long long ans;
    while(~scanf("%d %d",&n,&m))
    {
        memset(suma,0,sizeof suma);
        memset(sumb,0,sizeof sumb);
        la=lb=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&t);
            if(suma[t]==0)
                a[la++]=t;
            suma[t]++;
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d",&t);
            if(sumb[t]==0)
                b[lb++]=t;
            sumb[t]++;
        }
        ans=0;
        for(int i=0;i<la;i++)
            for(int j=0;j<lb;j++)
            {
                ans+=(long long)suma[a[i]]*sumb[b[j]]*(long long)(sqrt(abs(a[i]-b[j])));
            }
        printf("%lld\n",ans);
    }
    return 0;
}