CodeForces 620D Professor GukiZ and Two Arrays

Professor GukiZ andTwo Arrays

Time Limit:3000MS     MemoryLimit:262144KB     64bit IO Format:%I64d& %I64u

Description

Professor GukiZhas two arrays of integers, a and b. Professor wants to make thesum of the elements in the array as_a as closeas possible to the sum of the elements in the array bs_b. So he wants tominimize the value v = |s_a - s_b|.

In one operationprofessor can swap some element from the array a and some elementfrom the array b. For example if the array a is [5, 1, 3, 2, 4] and thearray b is [3, 3, 2] professorcan swap the element 5 from thearray a and the element 2 from thearray b and get the new array a[2, 1, 3, 2, 4] and thenew array b[3, 3, 5].

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

Input

The first linecontains integer n (1 ≤ n ≤ 2000) — the numberof elements in the array a.

The second linecontains n integers a_i ( - 10⁹ ≤ a_i ≤ 10⁹) — the elementsof the array a.

The third linecontains integer m (1 ≤ m ≤ 2000) — the numberof elements in the array b.

The fourth linecontains m integers b_j ( - 10⁹ ≤ b_j ≤ 10⁹) — the elementsof the array b.

Output

In the firstline print the minimal value v = |s_a - s_b| that canbe got with no more than two swaps.

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

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

If there areseveral optimal solutions print any of them. Print the swaps in order theprofessor did them.

Sample Input

Input

5

5 4 3 2 1

4

1 1 1 1

Output

1

2

1 1

4 2

Input

5

1 2 3 4 5

1

15

Output

0

0

Input

5

1 2 3 4 5

4

1 2 3 4

Output

1

1

3 1

题目不难，但是做了好多次，码力不够啊。

对于k=0或1时，直接暴力枚举。

k=2时，ans=min(abs(Suma-Sumb-2*(a[i]+a[j])+2*(b[k]+b[l])));

如果暴力枚举每一种需要O(n^4)，可以将b枚举出M*(M-1)的情况，再枚举a，对前面的情况进行二分，设T=Suma-Sumb-2*(a[i]+a[j])

要最小化ans，需找到比-T大的第一个数和比-T小的第一个数。

 

代码如下：

 

<pre name="code" class="cpp">#include<cstdio>#include<algorithm>#include<cstring>using namespace std;struct num{long long s;int no;};struct num2{long long s;int no1;int no2;};num a[2050];num b[2050];num2 d[2050*2050];int N,M;long long Suma=0,Sumb=0;long long ans;int tot=0;int xa[3];int xb[3];long long ABS(long long x){return x>=0?x:-x;}bool cmp(const num2 x,const num2 y){return x.s<y.s;}void swap(num &x,num &y){num z;z=x;x=y;y=z;}void getans1(){long long now=0;for (int i=1;i<=N;i++){for (int j=1;j<=M;j++){now=ABS(Suma-Sumb-2*a[i].s+2*b[j].s);if (now<ans){ans=now;tot=1;xa[1]=i;xb[1]=j;}}}}void getans2(){long long T;int l,r,mid;long long now=0;int NM=0;for (int i=1;i<=M;i++){for (int j=1;j<=M;j++){if (i!=j){NM++;d[NM].s=b[i].s+b[j].s;d[NM].no1=i;d[NM].no2=j;}}} sort(d+1,d+NM+1,cmp);for (int i=1;i<=N;i++){for (int j=1;j<=N;j++){if (i==j){continue;}T=Suma-Sumb-2*a[i].s-2*a[j].s;T=-T;l=1;r=NM;while (r-l>1){mid=(l+r)>>1;if (d[mid].s*2>T){r=mid;}else{l=mid;}}l=r-1;if (r<=NM){now=ABS(T-2*d[r].s);if (now<ans){ans=now;tot=2;xa[1]=i;xa[2]=j;xb[1]=d[r].no1;xb[2]=d[r].no2;}}if (l>=1){now=ABS(T-2*d[l].s);if (now<ans){ans=now;tot=2;xa[1]=i;xa[2]=j;xb[1]=d[l].no1;xb[2]=d[l].no2;}}}}}int main(){scanf("%d",&N);for (int i=1;i<=N;i++){scanf("%I64d",&a[i].s);a[i].no=i;Suma+=a[i].s;}scanf("%d",&M);for (int i=1;i<=M;i++){scanf("%I64d",&b[i].s);b[i].no=i;Sumb+=b[i].s;}ans=ABS(Suma-Sumb);tot=0;if (ans){getans1();}if (N>=2 && M>=2 && ans){getans2();}printf("%I64d\n",ans);printf("%d\n",tot);for (int i=1;i<=tot;i++){printf("%d %d\n",xa[i],xb[i]);}return 0;}