CC三元组题解

简单题直接上代码因为真的很简单

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#define itn int
#define reaD read
#define N 100005
#define mod 1000000007
#define int long long
using namespace std;
int T, p, q, r, a[N << 1], b[N << 1], c[N << 1], pos1, pos2, cnt1, cnt2, sum1, sum2; 
inline int read(){
	int x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if (c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	return x * w;
}
signed main(){
	T = read();
	while(T--){
		p = read(); q = read(); r = read();
		for(int i = 1; i <= p; i++)	a[i] = read();
		for(int i = 1; i <= q; i++)	b[i] = read();
		for(int i = 1; i <= r; i++)	c[i] = read();
		sort(a + 1, a + p + 1);
		sort(b + 1, b + q + 1);
		sort(c + 1, c + r + 1);
		long long ans = 0; 
		pos1 = pos2 = 1; cnt1 = cnt2 = 0; sum1 = sum2 = 0; 
		for(int i = 1; i <= q; i++)
		{
			while(pos1 <= p && a[pos1] <= b[i])
			{
				sum1 = 1ll * (sum1 + a[pos1]) % mod;
				cnt1++; pos1++; 
			}
			while(pos2 <= r && c[pos2] <= b[i])
			{
				sum2 = 1ll * (sum2 + c[pos2]) % mod;
				cnt2++; pos2++; 
			}
			ans = (1ll * ans + 1ll * cnt1 * cnt2 % mod * b[i] % mod * b[i] % mod) % mod;
			ans = (1ll * ans + 1ll * b[i] * (1ll * sum1 * cnt2 % mod + sum2 * cnt1 % mod) % mod) % mod;
			ans = (1ll * ans + 1ll * sum1 * sum2 % mod) % mod; 
		}
		printf("%lld
", ans); 
	}
	return 0;
}