F. Double Knapsack

题目连接：

http://www.codeforces.com/contest/618/problem/F

Description

You are given two multisets A and B. Each multiset has exactly n integers each between 1 and n inclusive. Multisets may contain multiple copies of the same number.

You would like to find a nonempty subset of A and a nonempty subset of B such that the sum of elements in these subsets are equal. Subsets are also multisets, i.e. they can contain elements with equal values.

If no solution exists, print - 1. Otherwise, print the indices of elements in any such subsets of A and B that have the same sum.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1 000 000) — the size of both multisets.

The second line contains n integers, denoting the elements of A. Each element will be between 1 and n inclusive.

The third line contains n integers, denoting the elements of B. Each element will be between 1 and n inclusive.

Output

If there is no solution, print a single integer - 1. Otherwise, your solution should be printed on four lines.

The first line should contain a single integer ka, the size of the corresponding subset of A. The second line should contain ka distinct integers, the indices of the subset of A.

The third line should contain a single integer kb, the size of the corresponding subset of B. The fourth line should contain kb distinct integers, the indices of the subset of B.

Elements in both sets are numbered from 1 to n. If there are multiple possible solutions, print any of them.

Sample Input

10

10 10 10 10 10 10 10 10 10 10

10 9 8 7 6 5 4 3 2 1

Sample Output

1

2

3

5 8 10

Hint

题意

给你a集合和b集合，两个集合里面都恰好有n个数，每个数都是在[1,n]范围内的数

然后你需要找到a集合的一个子集，b集合的一个子集，使得这两个子集的和相同

然后输出出来。

题解：

令Ai = a0+a1+...+ai,Bi = b0+b1+...+bi

其中A0 = 0,B0 = 0;

我们假设An<Bn,那么对于每个Ai,我们都可以找到一个最大的j，使得Ai-Bj>=0

显然有(n+1)对Ai-Bj,且0<=(Ai-Bj)<=n-1

根据鸽巢原理，显然可以找到两对Ai-Bj = Ai'-Bj'

所以只要输出这两队就好了

边扫边存下来就好了

然后这道题就结束了

代码

#include<bits/stdc++.h>

using namespace std;

const int maxn = 1e6+7;

const long long inf = 1e9;

long long a[maxn],b[maxn];

pair<int,int>d[maxn*2];

int main()

{

    int n;

    scanf("%d",&n);

    for(int i=1;i<=n;i++)

        scanf("%lld",&a[i]);

    for(int i=1;i<=n;i++)

        scanf("%lld",&b[i]);

    for(int i=0;i<maxn*2;i++)

        d[i]=make_pair(inf,inf);

    d[maxn]=make_pair(1,1);

    int p1=1,p2=1,now=0;

    while(1)

    {

        if(now<=0)now+=a[p1++];

        else now-=b[p2++];

        if(d[now+maxn].first!=inf)

        {

            printf("%d\n",p1-d[now+maxn].first);

            for(int i=d[now+maxn].first;i<p1;i++)

                printf("%d ",i);

            printf("\n");

            printf("%d\n",p2-d[now+maxn].second);

            for(int i=d[now+maxn].second;i<p2;i++)

                printf("%d ",i);

            printf("\n");

            return 0;

        }

        d[now+maxn]=make_pair(p1,p2);

    }

}

巴特西

Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) F. Double Knapsack 鸽巢原理构造

F. Double Knapsack

题目连接：

Description

Input

Output

Sample Input

Sample Output

Hint

题意

题解：

代码

最新文章

热门文章

巴特西

Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) F. Double Knapsack 鸽巢原理 构造

F. Double Knapsack

题目连接：

Description

Input

Output

Sample Input

Sample Output

Hint

题意

题解：

代码

最新文章

热门文章

Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) F. Double Knapsack 鸽巢原理构造