[数据结构与算法]C. Equal Sums

传送门

• 题意：题意：给 n 个数组，找出两个数组，这两个数组各删除一个数字后，使得两个数组的和相同.
• 题解：可以用一个map<int, pair<int, int> >的关键字存储第i个数组删除第j个值后总和，关键字对应的值为pair类型，pair分别储存对应的i，j值。
• 收获：学到如何判断上述类型的map是否被使用过。我是又开了一个map<int,int>数组，关键字与上述map相同。
  mp.count(sum - a[j]) != 0

my code：

const int N = 2e5, M = 1e6;
map<int, pii>id;
vector<int>v[N];
map<int, int>num;
int main()
{
	IOS;
	int k;
	cin >> k;
	int f = 0;
	for (int i = 0; i < k; i++)
	{
		int n; cin >> n;
		int x, sum = 0;
		for (int j = 0; j < n; j++)
		{
			cin >> x;
			sum += x;
			v[i].push_back(x);
		}

		for (int j = 0; j < n; j++)
		{
			if (num[sum - v[i][j]] == 0)
			{
				num[sum - v[i][j]]++;
				id[sum - v[i][j]].ft = i;
				id[sum - v[i][j]].sd = j;
			}
			else if(id[sum - v[i][j]].ft != i && !f)
			{
				f = 1;
				cout << "YES\n";
				cout << id[sum - v[i][j]].ft + 1 << " " << id[sum - v[i][j]].sd + 1 << endl;
				cout << i + 1 << " " << j + 1 << endl;
			}
		}
	}
	if(!f) cout << "NO\n";

	return 0;
}

others code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
LL k,n,sum,flag,a[N];
map <LL,pair<LL,LL>> mp;
int main(){
	ios :: sync_with_stdio(false);
	cin.tie(0);
	cin >> k;
	flag = false;
	for(int i = 1;i <= k;i++){
		cin >> n;
		sum = 0;
		for(int j = 1;j <= n;j++){
			cin >> a[j];
			sum += a[j];
		}
		if(flag) continue;
		for(int j = 1;j <= n;j++){
			if(mp.count(sum - a[j]) != 0){
				flag = true;
				cout << "YES" << endl << i << " " << j << endl << mp[sum - a[j]].first << " " << mp[sum -
				 a[j]].second << endl;
				break;
			}
		}
		for(int j = 1;j <= n;j++) mp[sum - a[j]] = {i,j};
	}
	if(!flag) cout << "NO" << endl;
	return 0;
}