Professor GukiZ and Two Arrays CodeForces - 620D （暴力枚举+二分搜索）

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本文详细解析了CodeForces-620D问题——通过最多两次元素交换使两个整数数组之和尽可能接近。介绍了输入输出格式、解决思路，并提供了高效的算法实现，包括不同交换次数下的时间复杂度分析。

Professor GukiZ and Two Arrays

CodeForces - 620D

Professor GukiZ has two arrays of integers, a and b. Professor wants to make the sum of the elements in the array a s_a as close as possible to the sum of the elements in the array b s_b. So he wants to minimize the value v = |s_a - s_b|.

In one operation professor can swap some element from the array a and some element from the array b. For example if the array a is [5, 1, 3, 2, 4] and the array b is [3, 3, 2] professor can swap the element 5 from the array a and the element 2 from the array b and get the new array a [2, 1, 3, 2, 4] and the new array b [3, 3, 5].

Professor doesn't want to make more than two swaps. Find the minimal value v and some sequence of no more than two swaps that will lead to the such value v. Professor makes swaps one by one, each new swap he makes with the new arrays a and b.

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of elements in the array a.

The second line contains n integers a_i ( - 10⁹ ≤ a_i ≤ 10⁹) — the elements of the array a.

The third line contains integer m (1 ≤ m ≤ 2000) — the number of elements in the array b.

The fourth line contains m integers b_j ( - 10⁹ ≤ b_j ≤ 10⁹) — the elements of the array b.

Output

In the first line print the minimal value v = |s_a - s_b| that can be got with no more than two swaps.

The second line should contain the number of swaps k (0 ≤ k ≤ 2).

Each of the next k lines should contain two integers x_p, y_p (1 ≤ x_p ≤ n, 1 ≤ y_p ≤ m) — the index of the element in the array a and the index of the element in the array b in the p-th swap.

If there are several optimal solutions print any of them. Print the swaps in order the professor did them.

Example

Input

Output

Input

Output

0
0

Input

Output

1
1
3 1

题意是有两个序列，可以不交换，交换一个数，交换两个数，使得两个数列和相减绝对值最小

不交换和交换一个比较简单直接枚举，时间分别为o(1)和o（m*n）

交换两个时，将第一个序列亮亮求和排序，枚举第二个序列的两个数，从第一个序列中二分查找最优解，时间o（m^2logn^2）

最后三个比一下就好了

题意是有两个序列，可以不交换，交换一个数，交换两个数，使得两个数列和相减绝对值最小

不交换和交换一个比较简单直接枚举，时间分别为o(1)和o（m*n）

交换两个时，将第一个序列亮亮求和排序，枚举第二个序列的两个数，从第一个序列中二分查找最优解，时间o（m^2logn^2）

最后三个比一下就好了

code：

    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<stdlib.h>
    #include<iostream>
    #include<algorithm>
    #include<functional>
    #include<queue>
    #include<vector>
    #include<set>
    #include<map>
    using namespace std;

    __int64 a[2005];
    __int64 b[2005];

    struct node
    {
        int l,r;
        __int64 h;
    }db[2000005];

    bool cmp(node a,node b)
    {
        if(a.h==b.h && a.l==b.l) return a.r<b.r;
        else if(a.h==b.h) return a.l<b.l;
        return a.h<b.h;
    }

    __int64 qfabs(__int64 a)
    {
        if(a<0) a=-a;
        return a;
    }

    int main()
    {
        int i,j,k;
        int n,m;
        __int64 suma,sumb;
        int ii,jj;
        int l,r,mid;
        __int64 ans0,ans1,ans2,ans;
        int a1,a2,b1,b2;

        suma=sumb=0;
        scanf("%d",&m);
        for(i=1;i<=m;i++)
        {
            scanf("%I64d",&a[i]);
            suma+=a[i];
        }
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%I64d",&b[i]);
            sumb+=b[i];
        }
        ans0=qfabs(suma-sumb);
        ans1=999999999;
        ans1=ans1*10000;
        for(i=1;i<=m;i++)
            for(j=1;j<=n;j++)
            {
                suma=suma-a[i]+b[j];
                sumb=sumb-b[j]+a[i];
                if(ans1>qfabs(suma-sumb))
                {
                    ans1=qfabs(suma-sumb);
                    ii=i;
                    jj=j;
                }
                suma=suma+a[i]-b[j];
                sumb=sumb+b[j]-a[i];
            }
        k=1;
        for(i=1;i<m;i++)
            for(j=i+1;j<=m;j++)
            {
                db[k].l=i;
                db[k].r=j;
                db[k].h=(a[i]+a[j])*2;
                k++;
            }
        sort(db+1,db+k,cmp);

        ans2=999999999;
        ans2=ans2*10000;
        if(k>1)//防止只有一个点
        {
            for(i=1;i<=n;i++)
                for(j=i+1;j<=n;j++)
                {
                    ans=suma+2*(b[i]+b[j])-sumb;
                    l=1;
                    r=k-1;
                    while(l<r-1)
                    {
                        mid=(l+r)/2;
                        if(db[mid].h==ans)
                        {
                            l=r=mid;
                            break;
                        }
                        if(db[mid].h>ans) r=mid;
                        if(db[mid].h<ans) l=mid;
                    }
                    if(qfabs(ans-db[l].h)>qfabs(db[r].h-ans) && qfabs(db[r].h-ans)<ans2)
                    {
                        b1=i;
                        b2=j;
                        a1=db[r].l;
                        a2=db[r].r;
                        ans2=qfabs(db[r].h-ans);
                    }
                    else if(qfabs(ans-db[l].h)<=qfabs(db[r].h-ans) && qfabs(db[l].h-ans)<ans2)
                    {
                        b1=i;
                        b2=j;
                        a1=db[l].l;
                        a2=db[l].r;
                        ans2=qfabs(db[l].h-ans);
                    }
                }
        }
        if(ans0<=ans1 && ans0<=ans2) printf("%d\n0\n",ans0);
        else if(ans1<=ans2) printf("%I64d\n1\n%d %d\n",ans1,ii,jj);
        else printf("%I64d\n2\n%d %d\n%d %d\n",ans2,a1,b1,a2,b2);
        return 0;
    }