CodeForces - 618F Double Knapsack

最新推荐文章于 2023-06-10 17:07:50 发布

王爷的大房子

最新推荐文章于 2023-06-10 17:07:50 发布

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原文链接：http://www.cnblogs.com/JYYHH/p/8927301.html

本文介绍了一种解决两个多重集合中寻找具有相同和的子集的问题。通过使用前缀和技巧，文章提出了一种高效的算法，能够在O(n)的时间复杂度内找到满足条件的子集，并附上了完整的C++实现代码。

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Discription

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 k_a, the size of the corresponding subset of A. The second line should contain k_a distinct integers, the indices of the subset of A.

The third line should contain a single integer k_b, the size of the corresponding subset of B. The fourth line should contain k_b 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.

Examples

Input

10
10 10 10 10 10 10 10 10 10 10
10 9 8 7 6 5 4 3 2 1

Output

Input

5
4 4 3 3 3
2 2 2 2 5

Output

2
2 3
2
3 5


    一道神构造。
    设sa[]为a[]的前缀和，sb[]为b的前缀和，对于每个0<=i<=n,我们找到一个最大的使得sb[j]<=sa[i]的j，这样每个sa[i]-sb[j]都是[0,n-1]的整数了(因为如果sa[i]-sb[j]>=n的话，j可以继续后移(a[],b[]中元素都<=n))
    所以至少会有一对 i和i'满足 sa[i]-sb[j] == sa[i']-sb[j']，直接输出就行了。。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1000005;
ll a[maxn],b[maxn];
int px[maxn],py[maxn],n;
inline int read(){
	int x=0; char ch=getchar();
	for(;!isdigit(ch);ch=getchar());
	for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x;
}
void W(int x){ if(x>=10) W(x/10); putchar(x%10+'0');}
int main(){
	n=read();
	for(int i=1;i<=n;i++) a[i]=read(),a[i]+=a[i-1];
	for(int i=1;i<=n;i++) b[i]=read(),b[i]+=b[i-1];
	for(int i=0,j=0,D;i<=n;i++){
		for(;j<n&&b[j+1]<=a[i];j++);
		D=a[i]-b[j];
		if(px[D]){
			W(i-px[D]+1),puts("");
			for(int o=px[D];o<=i;o++) W(o),putchar(' ');
			puts(""),W(j-py[D]+1),puts("");
			for(int o=py[D];o<=j;o++) W(o),putchar(' ');
			return 0;
		}
		px[D]=i+1,py[D]=j+1;
	}
	return 0;
}

转载于:https://www.cnblogs.com/JYYHH/p/8927301.html