【Educational Codeforces Round 6D】【暴力 SET二分】Professor GukiZ and Two Arrays

D. Professor GukiZ and Two Arrays

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

55 4 3 2 141 1 1 1

output

121 14 2

input

51 2 3 4 5115

output

00

input

51 2 3 4 541 2 3 4

output

113 1

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 2020, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int n,m;int a[N], b[N];int ansk; LL ansv;int P[6];void update(int k, LL v,int p1,int p2,int p3,int p4){v = abs(v);if (v < ansv){ansv = v;ansk = k;P[1] = p1;P[2] = p2;P[3] = p3;P[4] = p4;}}struct B{LL v;int p2, p4;B() {}B(LL v_, int p2_, int p4_) { v = v_; p2 = p2_; p4 = p4_; }bool operator < (const B&b)const{return v < b.v;}};set<B>sot;set<B>::iterator it;int main(){while(~scanf("%d",&n)){ansv = 1e18;LL suma = 0, sumb = 0;for (int i = 1; i <= n; ++i)scanf("%d", &a[i]), suma += a[i];scanf("%d", &m);for (int i = 1; i <= m; ++i)scanf("%d", &b[i]), sumb += b[i];update(0, suma - sumb,0,0,0,0);for (int i = 1; i <= n; ++i){for (int j = 1; j <= m; ++j){update(1,(suma-a[i]*2)-(sumb-b[j]*2),i,j,0,0);}}sot.clear();for (int i = 1; i < m; ++i){for (int j = i + 1; j <= m; ++j){sot.insert(B(sumb - b[i]*2 - b[j]*2, i, j));}}for (int i = 1; i <= n; ++i){for (int j = i + 1; j <= n; ++j){it = sot.lower_bound(B(suma - a[i] * 2 - a[j] * 2, -1, -1));if (it != sot.end())update(2, (suma - a[i] * 2 - a[j] * 2 - it->v), i, it->p2, j, it->p4);if (it != sot.begin()){--it;update(2, (suma - a[i] * 2 - a[j] * 2 - it->v), i, it->p2, j, it->p4);}}}printf("%lld\n", ansv);printf("%d\n", ansk);for (int i = 1; i <= ansk; ++i){printf("%d %d\n", P[i*2-1], P[i*2]);}}return 0;}/*【题意】给你两个数列a[],b[]（1<=a[],b[]<=1e9）长度分别为n,m（1<=n,m<=2000）我们可以选择(0~2)个pair，做两个数列的元素交换。问你如何交换，可以使得最后abs(suma-sumb)尽可能小【类型】暴力【分析】只是k==2的情况稍难处理。我们枚举第一个数列哪两个(i,j)和第二个数列做了交换。然后对于第二个数列，一样枚举两个交换位置x,y。我们发现交换后的abs()==abs( (suma-a[i]*2-a[j]*2)-(sumb-b[x]*2-b[y]*2) )于是我们把（sumb-b[x]*2-b[y]*2,x,y）存进set中然后二分(suma-a[i]*2-a[j]*2)，更新答案。【时间复杂度&&优化】O(n^2log(n^2))*/